دروس مادة الرياضيات للفصل الاول للشعب الادبية سنة ثالثة ثانوي

‫ﻓﻬﺭﺱ ﺍﻹﺭﺴﺎل ﺍﻷﻭل‬ ‫ﻴﺘﻀﻤﻥ ﻫﺫﺍ ﺍﻹﺭﺴﺎل ﺍﻟﻤﻭﺍﻀﻴﻊ ﺍﻟﺘﺎﻟﻴﺔ ‪:‬‬ ‫‪ −‬ﺍﻹﺴﺘﺩﻻل ﺒﺎﻟﺘﺭﺍﺠﻊ‬ ‫‪ -‬ﺍﻟﻤﺘﻨﺎﻟﻴﺎﺕ ﺍﻟﻌﺩﺩﻴﺔ‬‫‪ -‬ﺍﻟﻘﺴﻤﺔ ﺍﻹﻗﻠﻴﺩﻴﺔ ﻓﻲ ﻤﺠﻤﻭﻋﺔ ﺍﻷﻋﺩﺍﺩ ﺍﻟﺼﺤﻴﺤﺔ ‪ Z‬ﻭﺍﻟﻤﻭﺍﻓﻘـﺎﺕ‬ ‫ﻓﻲ ‪Z‬‬

‫‪ -1‬ﺍﻻﺴﺘﺩﻻل ﺒﺎﻟﺘﺭﺍﺠﻊ‬ ‫ﺍﻟﻜﻔﺎﺀﺍﺕ ﺍﻟﻤﺴﺘﻬﺩﻓﺔ ‪:‬‬‫‪ -‬ﺍﺴﺘﻌﻤﺎل ﻤﺒﺩﺃ ﺍﻻﺴﺘـﺩﻻل ﺒﺎﻟﺘﺭﺍﺠـﻊ ﻹﺜﺒﺎﺕ ﺼﺤﺔ ﺨﺎﺼﻴﺔ ﻤﻥ ﺃﺠل ﻜل ﻋﺩﺩ‬ ‫ﻁﺒﻴﻌﻲ ‪.n‬‬ ‫‪ -‬ﺍﻟﺘﻤﻴﻴﺯ ﺒﻴﻥ ﻤﺘﺘﺎﻟﻴﺔ ﻭ ﺤﺩﻫﺎ ﺍﻟﻌﺎﻡ‬ ‫‪ -‬ﺍﻟﺘﻌﺭﻑ ﻋﻠﻰ ﻤﺘﺘﺎﻟﻴﺔ ﺒﺎﻟﺘﺭﺍﺠﻊ‬ ‫‪ -‬ﺤﺴﺎﺏ ﺍﻟﺤﺩﻭﺩ ﺍﻷﻭﻟﻰ ﻟﻤﺘﺘﺎﻟﻴﺔ ﻤﻌﺭﻓﺔ ﺒﺎﻟﺘﺭﺍﺠﻊ‬ ‫‪ -‬ﺘﺤﺩﻴﺩ ﺍﺘﺠﺎﻩ ﺘﻐﻴﺭ ﻤﺘﺘﺎﻟﻴﺔ ﺤﺴﺎﺒﻴﺔ ﺃﻭ ﻫﻨﺩﺴﻴﺔ‬ ‫‪ -‬ﺍﺴﺘﻌﻤﺎل ﺍﻟﻤﺘﺘﺎﻟﻴﺎﺕ ﺍﻟﺤﺴﺎﺒﻴﺔ ﻭ ﺍﻟﻬﻨﺩﺴﻴﺔ ﻟﺤل ﻤﺸﻜﻼﺕ ﻤﻥ ﺍﻟﺤﻴﺎﺓ ﺍﻟﻴﻭﻤﻴﺔ‪.‬‬ ‫‪ -‬ﺍﻟﻤﺘﺘﺎﻟﻴﺎﺕ ﻤﻥ ﺍﻟﺸﻜل ‪:‬‬ ‫‪ Un+1 = a Un + b -‬ﻤﻊ ‪َ a ≠ 0‬و ‪b ≠ 0‬‬ ‫‪ -‬ﺤﺴﺎﺏ ﺍﻟﺤﺩ ﺍﻟﻌﺎﻡ ‪Un‬‬ ‫‪ -‬ﺤﺴﺎﺏ ‪ Sn‬ﻤﺠﻤﻭﻉ ‪ n‬ﺤﺩﹼﺍ ﻤﺘﺘﺎﺒﻌﺔ ﻟﻬﺫﻩ ﺍﻟﻤﺘﺘﺎﻟﻴﺔ‬ ‫‪ -‬ﺘﻌﻴﻴﻥ ﺍﺘﺠﺎﻩ ﺍﻟﺘﻐﻴﺭ‪.‬‬ ‫‪ -‬ﺤل ﻤﺸﻜﻼﺕ ﺘﺴﺘﻌﻤل ﻓﻴﻬﺎ ﻤﺘﺘﺎﻟﻴﺎﺕ ﻤﻥ ﺍﻟﺸﻜل‪:‬‬ ‫‪Un+1 = a Un +b‬‬ ‫ﺘﺼﻤﻴﻡ ﺍﻟﺩﺭﺱ‬ ‫‪ -‬ﺘﻌﺭﻴﻑ‬ ‫‪ -‬ﻤﺒﺩﺃ ﺍﻻﺴﺘﺩﻻل ﺒﺎﻟﺘﺭﺍﺠﻊ‬ ‫‪ -‬ﺘﻤﺎﺭﻴﻥ ﺤﻭل ﺍﻹﺴﺘﺩﻻل ﺒﺎﻟﺘﺭﺍﺠﻊ‬ ‫‪ -‬ﺤﻠﻭل ﺍﻟﺘﻤﺎﺭﻴﻥ‬

‫‪ -‬ﺘﻌﺭﻴﻑ ‪:‬‬‫ﺍﻻﺴﺘﺩﻻل ﺒﺎﻟﺘﺭﺍﺠﻊ ﻫﻭ ﻨﻤﻁ ﻤﻥ ﺃﻨﻤﺎﻁ ﺍﻟﺒﺭﻫﺎﻥ ﻴﺴﻤﺢ ﺒﺎﻟﺒﺭﻫﻨﺔ ﻋﻠﻰ ﺼﺤﺔ ﺨﺎﺼﻴﺔ ﺘﺘﻌﻠﻕ ﺒﻌﺩﺩ ﻁﺒﻴﻌﻲ‬ ‫ﻭﻴﻌﺘﻤﺩ ﻫﺫﺍ ﺍﻻﺴﺘﺩﻻل ﻋﻠﻰ ﺍﻟﻤﺒﺩﺃ ﺍﻟﺘﺎﻟﻲ ‪:‬‬ ‫‪ -‬ﻤﺒﺩﺃ ﺍﻻﺴﺘﺩﻻل ﺒﺎﻟﺘﺭﺍﺠﻊ ‪:‬‬ ‫)‪ P(n‬ﺨﺎﺼﻴﺔ ﺘﺘﻌﻠﻕ ﺒﺎﻟﻌﺩﺩ ﺍﻟﻁﺒﻴﻌﻲ ‪ n‬ﻭ ‪ n0‬ﻋﻨﺼﺭﺍ ﻤﻥ ‪.N‬‬ ‫ﻴﻤﻜﻨﻨﺎ ﺍﻟﺘﺄﻜﺩ ﻤﻥ ﺼﺤﺔ )‪ P(n‬ﻤﻥ ﺃﺠل ﻜل ‪ n≥n0‬ﺇﺫﺍ ﺘﺤﻘﻘﺎ ﺍﻟﺸﺭﻁﺎﻥ ﺍﻟﺘﺎﻟﻴﺎﻥ‪:‬‬ ‫‪ P(n0) -1‬ﺼﺤﻴﺤﺔ ) ﺍﻟﺨﺎﺼﻴﺔ ﻤﺤﻘﻘﺔ ﻤﻥ ﺃﺠل ﺍﻟﺭﺘﺒﺔ ‪( n0‬‬ ‫‪ -2‬ﻤﻥ ﺃﺠل ﻜل ‪ m≥n0‬ﻨﻔﺭﺽ ﺃﻥ )‪ P(m‬ﺼﺤﻴﺤﺔ ) ﻓﺭﻀﻴﺔ ﺍﻟﺘﺭﺍﺠﻊ(‬ ‫ﺇﺫﻥ )‪ P(m+1‬ﺼﺤﻴﺤﺔ‪.‬‬ ‫ﻭ ﺘﻜﻭﻥ ﻋﻨﺩﺌﺫ )‪ P(n‬ﺼﺤﻴﺤﺔ ﻤﻥ ﺃﺠل ﻜل ﻤﺭﺘﺒﺔ ‪ n‬ﺤﻴﺙ ‪m≥n0‬‬ ‫∗ﻤﻼﺤﻅﺔ‪: 1‬‬‫ﻟﻠﺘﻌﺒﻴﺭ ﻋﻠﻰ ﻜﻭﻥ ﺼﻭﺍﺏ )‪ P(m‬ﻴﺅﺩﻱ ﺇﻟﻰ ﺼـﻭﺍﺏ )‪ P(m+1‬ﻨﻘـﻭل ﺃﻥ ﺍﻟﺨﺎﺼـﻴﺔ ‪ P‬ﻭﺭﺍﺜﻴـﺔ)‬ ‫‪ ( héréditaire‬ﻤﻥ ﺃﺠل ﻤﺭﺘﺒﺔ ‪. n0‬‬ ‫*ﻤﻼﺤﻅﺔ‪: 2‬‬ ‫ﺍﻟﺸﻜل ﺍﻟﻤﻘﺎﺒل ﻴﻤﺜل ﺴﻠﻤﹰﺎ ﻨﺭﻴﺩ ﺍﻟﺼﻌﻭﺩ ﻓﻴﻪ ﺍﺒﺘﺩﺍﺀ ﻤﻥ ﺍﻟ ّﺩﺭﺠﺔ‬ ‫‪ n0‬ﺇﻟﻰ ﺍﻟﺩﺭﺠﺔ ‪. n+1‬‬ ‫‪N+1‬‬ ‫ﻴﻤﻜﻨﻨﺎ ﺍﺴﺘﻌﻤﺎل ﻤﺒﺩﺃ ﺍﻻﺴﺘﺩﻻل ﺒﺎﻟﺘﺭﺍﺠﻊ‬ ‫‪N‬‬ ‫ﻭ ﺫﻟﻙ ﺒﺎﻟﺼﻌﻭﺩ ﺃﻭﻻ ﺇﻟﻰ ﺍﻟﺩﺭﺠﺔ‪ n0‬ﺜﻡ‬ ‫ﺇﻟﻰ ﺍﻟﺩﺭﺠﺔ ‪ n1‬ﺍﻟﺘﻲ ﺘﻠﻴﻬﺎ ﻤﺒﺎﺸﺭﺓ ﻭ ﻫﻜﺫﺍ‬ ‫ﺇﻟﻰ ﺃﻥ ﻨﺼل ﺇﻟﻰ ﺍﻟﺩﺭﺠﺔ ﺫﺍﺕ ﺍﻟﻤﺭﺘﺒﺔ ‪N5‬‬‫)‪( n+1‬ﻭ ﻤﻨﻪ ﺼﻌﻭﺩﻨﺎ ﺍﻟﺴﻠﻡ ﻜﺎﻥ ‪N4‬‬ ‫‪N3‬‬‫‪N2‬‬ ‫ﺒﻁﺭﻴﻘﺔ ﺴﻠﻴﻤﺔ‪.‬‬‫ﻤﺜﺎل‪ :1‬ﺒﺭﻫﻥ ﺒﺎﻟﺘﺭﺍﺠﻊ ﻋﻠﻰ ﺼﺤﺔ ﺍﻟﻤﺴﺎﻭﺍﺓ ﺍﻟﺘﺎﻟﻴﺔ ﻭ ﺫﻟﻙ ﻤﻥ ﺃﺠل ﻜل ﻋﺩﺩ ﻁﺒﻴﻌﻲ ‪ n‬ﻏﻴﺭ ﻤﻌﺩ‪1‬ﻭﻤ‪N‬ﺎ‪.‬‬‫)‪N0 2 + 4 + 6 + …+ 2n = n(n+1‬‬

‫ﺍﻟﺠﻭﺍﺏ ‪:‬‬ ‫ﻨﻌﺘﺒﺭ ﺍﻟﺨﺎﺼﻴﺔ )‪ P(n‬ﺍﻟﺘﻲ ﺘﻌﺒﺭ ﻋﻥ ﺍﻟﻤﺴﺎﻭﺍﺓ‬ ‫)‪2 + 4 + 6 + …+ 2n = n(n+1‬‬ ‫ﺒﻤﺎ ﺃﻥ *‪ n ЄN‬ﺇﺫﻥ ﻨﺨﺘﺎﺭ ‪n0 = 1‬‬ ‫ﻨﺘﺤﻘﻕ ﻤﻥ ﺼﺤﺔ ) ‪P(n0‬‬ ‫‪E1 = 2 + 4 + 6 + …. + 2n‬‬ ‫ﻨﻀﻊ‬ ‫)‪E2 = n(n+1‬‬ ‫َﻭ‬ ‫ﻤﻥ ﺃﺠل ‪ n0 = 1‬ﻟﺩﻴﻨﺎ ‪E1 = 2(1) = 2‬‬ ‫‪ E2 = 1(1+1) = 2‬ﺇﺫﻥ )‪ P(n0‬ﺼﺤﻴﺤﺔ‪.‬‬‫‪ -‬ﻨﻔﺭﺽ ﺼﺤﺔ )‪ P(n‬ﻤﻥ ﺃﺠل ﺍﻟﻤﺭﺘﺒﺔ ‪ m‬ﺤﻴﺙ ‪ m≥n0‬ﻭ ﻨﺒﺭﻫﻥ ﺼﺤﺔ )‪P(m+1‬‬ ‫ﻓﺭﻀﻴﺔ ﺍﻟﺘﺭﺍﺠﻊ‪ P(m) :‬ﺼﺤﻴﺤﺔ ﺃﻱ‬ ‫)‪2 + 4 + 6 + …..+ 2m = m(m+1‬‬ ‫ﻨﺒﺭﻫﻥ ﺼﺤﺔ )‪ P(m+1‬ﺃﻱ‪:‬‬‫)‪2 + 4 + 6 + …..+ 2m + 2(m+1) = (m+1)(m+2‬‬ ‫ﻭ ﻤﻨﻪ )‪E1= 2 + 4 + 6 + …..+ 2m + 2(m+1‬‬ ‫ﻤﻥ ﺼﻭﺍﺏ )‪P(m‬‬ ‫)‪E1= m(m+1) + 2(m+1‬‬ ‫)‪= (m+1)(m+2‬‬ ‫ﺇﺫﻥ ‪E1 = E2‬‬ ‫ﻭ ﻤﻨﻪ )‪ P(m+1‬ﺼﺤﻴﺤﺔ‬ ‫ﻭ ﻋﻠﻴﻪ ﻤﻥ ﺃﺠل ﻜل ‪ n‬ﻤﻥ * ‪ P(n) , N‬ﺼﺤﻴﺤﺔ‪.‬‬‫ﻤﺜﺎل‪ : 2‬ﺒﺭﻫﻥ ﺒﺎﻟﺘﺭﺍﺠﻊ ﻤﻥ ﺃﺠل ﻜل ﻋﺩﺩ ﻁﺒﻴﻌﻲ ‪ n‬ﺼﺤﺔ ﺍﻟﺨﺎﺼﻴﺔ)‪ P(n‬ﺍﻟﻤﻌﺭﻓﺔ ﺒﺎﻟﻤﺴﺎﻭﺍﺓ‪:‬‬‫‪0²‬‬ ‫‪+‬‬ ‫‪2²‬‬ ‫‪+‬‬ ‫‪4²‬‬ ‫‪+‬‬ ‫‪6²‬‬ ‫‪+‬‬ ‫‪…+‬‬ ‫‪(2n)²‬‬ ‫=‬ ‫‪2‬‬ ‫)‪n(n+1)(2n+1‬‬ ‫‪3‬‬ ‫ﺍﻷﺠﻭﺒﺔ ‪:‬‬ ‫ﺒﻤﺎ ﺃﻥ ‪ n ЄN‬ﻨﺄﺨﺫ ‪. n0 = 0‬‬ ‫‪ x‬ﻨﺘﺤﻘﻕ ﻤﻥ ﺼﺤﺔ )‪. P(n0‬‬ ‫ﺍﻟﻁﺭﻑ ﺍﻷﻭل ﻤﻥ ﺍﻟﻤﺴﺎﻭﺍﺓ‪E1 = 0² = 0 :‬‬ ‫=‪E1‬‬ ‫‪2‬‬ ‫ﺍﻟﻤﺴﺎﻭﺍﺓ‪(0)(0+1)(2*0+1)=0:‬‬ ‫ﻤﻥ‬ ‫ﺍﻟﺜﺎﻨﻲ‬ ‫ﺍﻟﻁﺭﻑ‬ ‫‪3‬‬ ‫ﺇﺫﻥ ‪ E1 = E2‬ﻭ ﻋﻠﻴﻪ ﻓﺈﻥ )‪ P(n0‬ﺼﺤﻴﺤﺔ‪.‬‬

‫‪ x‬ﻨﻔﺭﺽ ﺼﺤﺔ )‪ P(n‬ﻤﻥ ﺃﺠل ﺍﻟﻤﺭﺘﺒﺔ ‪ m≥n0‬ﺃﻱ ‪:‬‬ ‫‪0² + 2² + 4² + …+ (2m)²‬‬ ‫=‬ ‫‪2‬‬ ‫)‪m(m+1)(2m+1‬‬ ‫‪3‬‬ ‫ﻭ ﻨﺒﺭﻫﻥ ﺼﺤﺔ )‪ P(m+1‬ﺃﻱ‬ ‫‪0²‬‬ ‫‪+‬‬ ‫‪2²‬‬ ‫‪+‬‬ ‫‪…+‬‬ ‫‪(2m)²‬‬ ‫‪+‬‬ ‫‪[2(m+1)]²‬‬ ‫=‬ ‫‪2‬‬ ‫)‪(m+1)(m+2)(2m+3‬‬ ‫‪3‬‬‫ﻟﺩﻴﻨﺎ ‪E1= 0² + 2² + 4² + …+ (2m)² +[2(m+1)]²‬‬ ‫ﻤﻥ ﺼﻭﺍﺏ )‪P(m‬‬ ‫‪E1‬‬ ‫=‬ ‫‪2‬‬ ‫‪m(m+1)(2m+1) +4 (m+1)2‬‬ ‫‪3‬‬‫=‬ ‫‪2‬‬ ‫@)‪(m+1)>m(2m+1)+6(m+1‬‬ ‫‪3‬‬ ‫ﻭ ﺫﻟﻙ ﺒﺘﻭﺤﻴﺩ ﺍﻟﻤﻘﺎﻤﺎﺕ ﻭ ﺇﺨﺭﺍﺝ ﺍﻟﻌﺎﻤل ﺍﻟﻤﺸﺘﺭﻙ‬ ‫= ‪E1‬‬ ‫‪2‬‬ ‫>)‪(m+1‬‬ ‫‪2m2+7m‬‬ ‫)‪+6‬‬ ‫ﺇﺫﻥ‪:‬‬ ‫‪3‬‬ ‫=‪E1‬‬ ‫‪2‬‬ ‫()‪(m+1‬‬ ‫)‪m+2)(2m+3‬‬ ‫‪3‬‬ ‫‪E1 = E2‬‬ ‫ﺇﺫﻥ‪:‬‬‫ﻭ ﻤﻨﻪ )‪ P(m+1‬ﺼﺤﻴﺤﺔﻭ ﻨﺴﺘﺨﻠﺹ ﺃﻨﻪ ﻤﻥ ﺃﺠل ﻜل ‪ n‬ﻤﻥ ‪ N‬ﻓﺈﻥ )‪ P(n‬ﺼﺤﻴﺤﺔ‪.‬‬ ‫‪ -‬ﺘﻤﺎﺭﻴﻥ ﺤﻭل ﺍﻹﺴﺘﺩﻻل ﺒﺎﻟﺘﺭﺍﺠﻊ ‪:‬‬ ‫♦ﺍﻟﺘﻤﺭﻴﻥ ‪: 1‬‬‫ﺒﺭﻫﻥ ﺒﺎﻟﺘﺭﺍﺠﻊ ﻤﻥ ﺃﺠل ﻜل ﻋﺩﺩ ﻁﺒﻴﻌﻲ ‪ n‬ﻏﻴﺭ ﻤﻌﺩﻭﻤﺎ ﺼﺤﺔ ﺍﻟﻤﺴﺎﻭﺍﺓ ‪:‬‬ ‫‪12 + 22 + 32 +…+ n2‬‬ ‫=‬ ‫(‬ ‫‪1‬‬ ‫)‬ ‫)‪n (2n2+3n+1‬‬ ‫‪6‬‬ ‫♦ﺍﻟﺘﻤﺭﻴﻥ ‪: 2‬‬ ‫ﺒﺭﻫﻥ ﻤﻥ ﺃﺠل ﻜل ‪ n‬ﻤﻥ ‪ N‬ﻭﺒﺎﻟﺘﺭﺍﺠﻊ ﺼﺤﺔ ﺍﻟﻤﺴﺎﻭﺍﺓ ‪:‬‬ ‫‪03‬‬ ‫‪+‬‬ ‫‪23‬‬ ‫‪+‬‬ ‫‪43‬‬ ‫‪+‬‬ ‫…‬ ‫‪+(2n)3‬‬ ‫(=‬ ‫‪1‬‬ ‫)‬ ‫‪n2 (n+1)2‬‬ ‫‪4‬‬

‫♦ﺍﻟﺘﻤﺭﻴﻥ ‪: 3‬‬ ‫ﺒﺭﻫﻥ ﺒﺎﻟﺘﺭﺍﺠﻊ ﻤﻥ ﺃﺠل ﻜل ‪ n‬ﻤﻥ *‪: N‬‬ ‫‪1‬‬ ‫‪+‬‬ ‫(‬ ‫‪1‬‬ ‫‪+‬‬ ‫‪1‬‬ ‫‪+…+‬‬ ‫‪1‬‬ ‫)‪1‬‬ ‫=‬ ‫‪n‬‬‫)‪(1u2‬‬ ‫)‪2u3‬‬ ‫)‪(3u4‬‬ ‫ ‪n(n‬‬ ‫‪n 1‬‬ ‫ﺤﻠﻭل ﺍﻟﺘﻤﺎﺭﻴﻥ‬ ‫• ﺍﻟﺘﻤﺭﻴﻥ ﺍﻷﻭل ‪:‬‬ ‫ﻨﺴﻤﻲ )‪ p(n‬ﺍﻟﺨﺎﺼﻴﺔ ﺍﻟﻤﺘﻌﻠﻘﺔ ﺒﺎﻟﻁﺒﻴﻌﻲ ‪ n‬ﻭﺍﻟﻤﻌﺭﻓﺔ ﺒﺎﻟﻤﺴﺎﻭﺍﺓ ‪:‬‬ ‫‪12‬‬ ‫‪+‬‬ ‫‪22 +‬‬ ‫‪32‬‬ ‫‪+‬‬ ‫‪…+‬‬ ‫= ‪n2‬‬ ‫‪1‬‬ ‫‪n(2‬‬ ‫‪n2‬‬ ‫‪3n‬‬ ‫)‪1‬‬ ‫‪6‬‬ ‫ﻨﺴﻤﻲ ‪E1 = 12 + 22 + 32 +…n2 :‬‬ ‫ﻭ‬ ‫= ‪E2‬‬ ‫‪1‬‬ ‫‪n‬‬ ‫(‬ ‫‪2n‬‬ ‫‪2‬‬ ‫)‪ 3 n 1‬‬ ‫‪6‬‬ ‫‪ (1‬ﻨﺘﺤﻘﻕ ﻤﻥ ﺼﺤﺔ )‪ p(n0‬ﺤﻴﺙ ‪n0 = 1‬‬ ‫‪E1 = 12 = 1‬‬ ‫‪E2‬‬ ‫=‬ ‫(‬ ‫‪1‬‬ ‫)‬ ‫)‪(1‬‬ ‫‪(2(1)2‬‬ ‫‪+‬‬ ‫)‪3(1‬‬ ‫‪+‬‬ ‫)‪1‬‬ ‫(=‬ ‫‪1‬‬ ‫)‬ ‫‪(2‬‬ ‫‪+‬‬ ‫‪3‬‬ ‫‪+‬‬ ‫)‪1‬‬ ‫‪6‬‬ ‫‪6‬‬ ‫ﺇﺫﻥ )‪ p(n0‬ﺼﺤﻴﺤﺔ‪.‬‬ ‫= ‪E2‬‬ ‫‪6‬‬ ‫‪=1‬‬ ‫‪6‬‬ ‫‪ (2‬ﻨﻔﺭﺽ ﺃﻥ )‪ p(n‬ﺼﺤﻴﺤﺔ ﻭﻨﺒﺭﻫﻥ ﺃﻥ )‪ p(n0‬ﺼﺤﻴﺤﺔ ﻤﻥ ﺃﺠل ﻜل ‪n tn0‬‬ ‫ﺃﻱ ‪:‬‬ ‫‪12‬‬ ‫‪+‬‬ ‫‪22‬‬ ‫‪+‬‬ ‫‪32‬‬ ‫‪+‬‬ ‫‪…+‬‬ ‫‪n2‬‬ ‫=‬ ‫(‬ ‫‪1‬‬ ‫)‬ ‫‪n(2n2‬‬ ‫‪+‬‬ ‫‪3n‬‬ ‫‪+‬‬ ‫)‪1‬‬ ‫‪6‬‬ ‫ﻭﻨﺒﺭﻫﻥ ﺃﻥ ‪:‬‬ ‫‪12+…+‬‬ ‫(=‪(n2)+(n+1)2‬‬ ‫‪1‬‬ ‫‪)(n+1)[2(n+1)2+3(n+1)+‬‬ ‫]‪1‬‬ ‫‪6‬‬

‫‪12‬‬ ‫‪+‬‬ ‫‪22‬‬ ‫‪+‬‬ ‫‪..+‬‬ ‫‪(n2)+(n+1)2‬‬ ‫=‬ ‫(‬ ‫‪1‬‬ ‫)‬ ‫)‪(n+1‬‬ ‫‪[2n2‬‬ ‫‪+‬‬ ‫‪7n‬‬ ‫‪+‬‬ ‫]‪6‬‬ ‫‪6‬‬ ‫ﻟﺩﻴﻨﺎ‪:‬‬ ‫‪E1=12+22+32+. . .+n2 +(n+1)2‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫(=‪E1‬‬ ‫‪1‬‬ ‫)‬ ‫‪n(2n2‬‬ ‫‪+‬‬ ‫‪3n‬‬ ‫‪+‬‬ ‫‪1)+(n+1)2‬‬ ‫‪6‬‬ ‫‪E1‬‬ ‫=‬ ‫‪1‬‬ ‫‪n(2n+1)(n+1)+(n+1)2‬‬ ‫‪6‬‬ ‫‪1‬‬ ‫@)‪E1 = 6 (n+1)>n(2n+1)+6(n+1‬‬ ‫‪E1‬‬ ‫=‬ ‫‪1‬‬ ‫)‪(n+1‬‬ ‫@‪)>2n2+7n+6‬‬ ‫‪6‬‬ ‫ﻻﺤﻅ ﺃﻥ ‪ E1= E2‬ﺍﺫﻥ )‪p(n+1‬ﺼﺤﻴﺤﺔ ﻭﻤﻨﻪ ﻤﻥ ﺃﺠل ﻜل ‪ n‬ﻤﻥ*‪ p(n) . N‬ﺼﺤﻴﺤﺔ‪.‬‬ ‫• ﺍﻟﺘﻤﺭﻴﻥ ﺍﻟﺜﺎﻨﻲ ‪ :‬ﻴﺤل ﺒﻨﻔﺱ ﺍﻟﻁﺭﻴﻘﺔ‬ ‫• ﺍﻟﺘﻤﺭﻴﻥ ﺍﻟﺜﺎﻟﺙ ‪:‬‬ ‫‪1‬‬ ‫‪+‬‬ ‫(‬ ‫‪1‬‬ ‫‪+‬‬ ‫‪1‬‬ ‫‪+…+‬‬ ‫‪n‬‬ ‫(‬ ‫‪1‬‬ ‫=‬ ‫‪n‬‬‫)‪(1u2‬‬ ‫)‪2u3‬‬ ‫)‪(3u4‬‬ ‫)‪n 1‬‬ ‫‪n 1‬‬ ‫ﻨﻀﻊ )‪ p(n‬ﺍﻟﺨﺎﺼﻴﺔ ﺍﻟﺘﻲ ﺘﺤﻘﻕ‬ ‫‪1‬‬ ‫‪+‬‬ ‫(‬ ‫‪1‬‬ ‫‪+‬‬ ‫‪1‬‬ ‫‪+…+‬‬ ‫‪n‬‬ ‫(‬ ‫‪1‬‬ ‫=‬ ‫‪n‬‬‫)‪(1u2‬‬ ‫)‪2u3‬‬ ‫)‪(3u4‬‬ ‫)‪n 1‬‬ ‫‪n 1‬‬ ‫ﻨﺘﺤﻘﻕ ﻤﻥ ﺼﺤﺔ )‪p(n0‬ﺤﻴﺙ ‪n0=1‬‬ ‫ﺍﻟﻁﺭﻑ ﺍﻷﻭل‪:‬‬ ‫‪11‬‬‫‪E1= (1u2) = 2‬‬

‫ﺍﻟﻁﺭﻑ ﺍﻟﺜﺎﻨﻲ‪:‬‬ ‫‪11‬‬ ‫‪E2= (1  1) = 2‬‬ ‫)‪ P(n0‬ﺼﺤﻴﺤﺔ‬ ‫ﻨﻔﺭﺽ ﺼﺤﺔ )‪ p(n‬ﻭﻨﺒﺭﻫﻥ ﺼﺤﺔ )‪p(n+1‬ﻭﺫﻟﻙ ﻤﻥ ﺍﺠل ‪n≥n0‬‬ ‫‪1‬‬ ‫‪+‬‬ ‫‪1‬‬ ‫‪+…+‬‬ ‫‪n‬‬ ‫(‬ ‫‪1‬‬ ‫=‬ ‫‪n‬‬ ‫)‪(1u2‬‬ ‫)‪( 2u3‬‬ ‫)‪n 1‬‬ ‫‪n 1‬‬ ‫ﻨﺒﺭﻫﻥ ﺃﻥ ‪:‬‬ ‫‪1‬‬ ‫‪+‬‬ ‫(‬ ‫‪1‬‬ ‫‪+….+‬‬ ‫‪n‬‬ ‫(‬ ‫‪1‬‬ ‫‪+‬‬ ‫‪(n‬‬ ‫‬ ‫‪1‬‬ ‫‬ ‫)‪2‬‬ ‫=‬ ‫‪(n‬‬ ‫‬ ‫)‪1‬‬‫)‪(1u2‬‬ ‫)‪2u3‬‬ ‫)‪n 1‬‬ ‫‪1)(n‬‬ ‫‪(n‬‬ ‫‬ ‫)‪2‬‬ ‫ﺘﻭﺤﻴﺩ ﺍﻟﻤﻘﺎﻤﺎﺕ‬ ‫ﻭﺒﻌﺩ‬ ‫= ‪E1‬‬ ‫‪n1‬‬ ‫)‪2‬‬ ‫ ‪n1 + (n  1)(n‬‬ ‫‪ E1 = n(n2)1‬ﻭﻤﻨﻪ‬ ‫) ‪( n 1)( n  2‬‬ ‫‪n2 2n1‬‬ ‫‪( n 1) 2‬‬ ‫)‪E1 = (n1)(n2) = (n2)(n1‬‬ ‫ﺒﻌﺩ ﺍﻹﺨﺘﺯﺍل )‪ (n+1‬ﺒﺴﻁﺎ ﻭﻤﻘﺎﻤﺎ ﻨﺠﺩ ‪:‬‬ ‫= ‪E1‬‬ ‫‪n 1‬‬ ‫‪n2‬‬ ‫ﺍﺫﻥ ‪ E1 = E2‬ﻭﻤﻨﻪ )‪ p(n+1‬ﺼﺤﻴﺤﺔ‬ ‫ﻭﻤﻨﻪ ﻤﻥ ﺃﺠل ﻜل ‪ n‬ﻤﻥ *‪ N‬ﻓﺈﻥ )‪ p(n‬ﻤﺤﻘﻘﺔ‬

‫‪ -2‬ﺍﻟﻤﺘﺘﺎﻟﻴﺎﺕ ﺍﻟﻌﺩﺩﻴﺔ‬ ‫‪ -1‬ﺘﻌﺭﻴﻑ ﻤﺘﺘﺎﻟﻴﺔ‬ ‫‪ -2‬ﺍﻟﺘﻌﺭﻑ ﻋﻠﻰ ﻤﺘﺘﺎﻟﻴﺔ ﺒﺎﻟﺘﺭﺍﺠﻊ‬ ‫‪ -3‬ﺇﺘﺠﺎﻩ ﺘﻐﻴﺭ ﻤﺘﺘﺎﻟﻴﺔ \"ﺭﺘﺎﺒﺔ ﻤﺘﺘﺎﻟﻴﺔ\"‬ ‫‪ - 4‬ﺍﻟﻤﺘﺘﺎﻟﻴﺎﺕ ﺍﻟﺤﺴﺎﺒﻴﺔ ﻭﺍﻟﻬﻨﺩﺴﻴﺔ‬ ‫‪ -5‬ﺇﺘﺠﺎﻩ ﺘﻐﻴﺭ ﻤﺘﺘﺎﻟﻴﺔ ﺤﺴﺎﺒﻴﺔ ﻭﻤﺘﺘﺎﻟﻴﺔ ﻫﻨﺩﺴﻴﺔ‬‫‪ -6‬ﺇﺴﺘﻌﻤﺎل ﺍﻟﻤﺘﺘﺎﻟﻴﺎﺕ ﺍﻟﺤﺴﺎﺒﻴﺔ ﻭ ﺍﻟﻬﻨﺩﺴﻴﺔ ﻟﺤل ﻤﺸﻜﻼﺕ ﻤﻥ ﺍﻟﺤﻴﺎﺓ‬ ‫ﺍﻟﻴﻭﻤﻴﺔ‬ ‫‪ -7‬ﺍﻟﻤﺘﺘﺎﻟﻴﺎﺕ ﻤﻥ ﺍﻟﺸﻜل ‪ U n1 aU n  b :‬ﻤﻊ ‪ a≠0‬ﻭ ‪b≠0‬‬ ‫‪ -‬ﺘﻤﺎﺭﻴﻥ ﻭﻤﺸﻜﻼﺕ ﺤﻭل ﺍﻟﻤﺘﺘﺎﻟﻴﺎﺕ ﺍﻟﻌﺩﺩﻴﺔ‬ ‫‪ -‬ﺤـﻠﻭل ﺍﻟﺘﻤﺎﺭﻴﻥ‬

‫‪ -1‬ﺘﻌﺭﻴﻑ ﻤﺘﺘﺎﻟﻴﺔ ‪:‬‬ ‫ﺍﻟﻤﺘﺘﺎﻟﻴﺔ ﺍﻟﻌﺩﺩﻴﺔ ﻫﻲ ﺩﺍﻟﺔ ﻤﺠﻤﻭﻋﺔ ﺘﻌﺭﻴﻔﻬﺎ ‪ N‬ﺃﻭ ﺠﺯﺀ ﻤﻥ ‪N‬‬ ‫ﻤﺜل‪ N * :‬ﺃﻭ`‪N* ^1‬‬ ‫*ﻤﻼﺤﻅﺔ‪ : 1‬ﻴﺠﺏ ﺃﻥ ﻨﻔﺭﻕ ﺒﻴﻥ ﺍﻟﻌﺒﺎﺭﺘﻴﻥ )‪ (Un‬ﻭ ‪Un‬‬ ‫ﺤﻴﺙ )‪ (Un‬ﻴﺭﻤﺯ ﺒﻪ ﺇﻟﻰ ﺍﻟﻤﺘﺘﺎﻟﻴﺔ ﻭ ‪ Un‬ﻟﻌﺒﺎﺭﺓ ﺍﻟﺤﺩ ﺍﻟﻌﺎﻡ ﻟﻠﻤﺘﺘﺎﻟﻴﺔ )‪.(Un‬‬ ‫*ﻤﻼﺤﻅﺔ ‪ : 2‬ﻤﻥ ﺍﻟﺘﻌﺭﻴﻑ ﺃﻋﻼﻩ ﺍﻟﻤﺘﺘﺎﻟﻴﺔ ﺍﻟﻤﻌﺭﻓﺔ ﺍﺒﺘﺩﺍﺀﺍ ﻤﻥ ﺍﻟﻤﺭﺘﺒﺔ ‪ n0‬ﻓﻤﺜﻼ‬ ‫‪1‬‬ ‫‪ Un = n‬ﻫﻭ ﺍﻟﺤﺩ ﺍﻟﻌﺎﻡ ﻟﻠﻤﺘﺘﺎﻟﻴﺔ )‪ (Un‬ﺍﻟﻤﻌﺭﻓﺔ ﻋﻠﻰ *‪.N‬‬ ‫ﻭﺃﻤﺎ ‪ Vn = n  2‬ﻓﻴﻤﺜل ﺍﻟﺤﺩ ﺍﻟﻌﺎﻡ ﻟﻠﻤﺘﺘﺎﻟﻴﺔ)‪ (Vn‬ﺍﻟﻤﻌﺭﻓﺔ ﻤﻥ ﺃﺠل ‪n. ≥ 2‬‬‫ﻭﻤﻨﻪ ﻤﺠﻤﻭﻋﺔ ﺘﻌﺭﻴﻑ ﻤﺘﺘﺎﻟﻴﺔ ﻫﻲ ﻤﻥ ﺍﻟﺸﻜل ‪ [n0 , ∞[ :‬ﺤﻴﺙ ‪.n0 є N‬‬‫‪ -2‬ﺍﻟﺘﻌﺭﻑ ﻋﻠﻰ ﻤﺘﺘﺎﻟﻴﺔ ﺒﺎﻟﺘﺭﺍﺠﻊ ‪:‬‬‫ﻨﺴﻤﻲ ﻤﺘﺘﺎﻟﻴﺔ ﺘﺭﺍﺠﻌﻴﺔ ﻜل ﻤﺘﺘﺎﻟﻴﺔ ﻤﻌﺭﻓﺔ ﻋﻠﻰ ‪ N‬ﺒﻌﻼﻤﺔ ﺘﺴﻤﺢ ﺒﺘﻌﻴﻴﻥ ﻜل ﺤ ّﺩ ﻤﻨﻬﺎ ﺍﻨﻁﻼﻗﺎ ﻤﻥ ﺤـﺩﻭﺩ‬ ‫ﺴﺒﻕ ﻤﻌﺭﻓﺘﻬﺎ‪.‬‬‫ﻜﺫﻟﻙ ﻴﻤﻜﻨﻨﺎ ﺃﻥ ﻨﻌﺭﻑ ﻤﺘﺘﺎﻟﻴﺔ ﺘﺭﺍﺠﻌﻴﺔ ﻜل ﻤﺘﺘﺎﻟﻴﺔ ‪ U‬ﺍﻟﻤﻌﺭﻓﺔ ﺒﺤﺩﻫﺎ ﺍﻷﻭل ‪ Un0‬ﻭﺒﺎﻟﻌﻼﻗـﺔ‬‫)‪ Un+1= f(Un‬ﺤﻴﺙ ‪ f‬ﺩﺍﻟﺔ ﻤﻌﺭﻓﺔ ﻋﻠﻰ ‪.N‬‬‫ﺍﻟﺩﺍﻟﺔ ‪ f‬ﺘﺴﻤﻰ ﺍﻟﺩﺍﻟﺔ ﺍﻟﻤﺭﻓﻘﺔ ﺒﺎﻟﻤﺘﺘﺎﻟﻴﺔ ‪.U‬‬‫ﻤﺜﺎل‪ (Un) :1‬ﻤﺘﺘﺎﻟﻴﺔ ﻤﻌﺭﻓﺔ ﻋﻠﻰ ‪ N‬ﺒﺎﻟﻌﻼﻗﺔ ﺍﻟﺘﺭﺍﺠﻌﻴﺔ‬ ‫‪U0=3‬‬‫ﻤﻥ ﺃﺠل ﻜل ﻁﺒﻴﻌﻲ ‪ n‬ﻤﻥ ‪N‬‬ ‫‪Un+1= 4Un-6‬‬‫‪-(1‬ﺃﺤﺴﺏ‪U3،U2،U1 :‬‬‫‪-(2‬ﻨﻌﺘﺒﺭ ﺍﻟﻤﺘﺘﺎﻟﻴﺔ )‪ (Vn‬ﺍﻟﻤﻌﺭﻓﺔ ﻤﻥ ﺃﺠل ﻜل ﻤﻥ ﺃﺠل ﻜل ‪ n‬ﻤﻥ ‪N‬‬‫ﺒﻌﺒﺎﺭﺓ ﺤﺩﻫﺎ ﺍﻟﻌﺎﻡ‪Vn = Un -2 :‬‬‫* ﺍﺤﺴﺏ ‪V3،V2،V1،V0‬‬‫* ﺒﺘﺨﻤﻴﻥ ﺍﺤﺴﺏ ‪ V8‬ﺜﻡ ﺃﻋﻁ ﻋﺒﺎﺭﺓ ‪ Vn‬ﺒﺩﻻﻟﺔ ‪n‬‬

‫ﺍﻷﺠﻭﺒﺔ ‪:‬‬ ‫‪U0 = 3‬‬ ‫‪Un+1 = 4Un-6‬‬ ‫‪ -1‬ﺤﺴﺎﺏ‪U3،U2،U1 :‬‬‫‪U1 = 4U0-6 = 4(3)-6 = 6‬‬‫‪U2 = 4U1-6 = 4(6)-6 = 24-6=18‬‬‫‪U3 = 4U2-6 = 4(18)-6 = 66‬‬ ‫‪ -2‬ﺤﺴﺎﺏ ٍ‪V3،V2،V1‬‬ ‫ﻟﺩﻴﻨﺎ ‪Vn = Un-2‬‬ ‫‪V0 = U0-2 = 3-2 = 1‬‬ ‫‪V1 = U1-2 = 6-2 = 4‬‬‫‪V2 = U2-2 = 18-2=16‬‬‫‪V3 = U3-2 = 66-2 = 64‬‬ ‫*ﺤﺴﺎﺏ ‪ V8‬ﺒﺘﺨﻤﻴﻥ‪:‬‬ ‫ﻟﺩﻴﻨﺎ ‪V3 = 43 ، V2 = 42 ، V1 = 41 ،V0 = 1= 40‬‬ ‫ﺇﺫﻥ ‪ V8 = 48‬ﺃﻱ ‪V8 = 65536‬‬ ‫ﻭﻤﻨﻪ ‪Vn = 4n‬‬‫ﻤﺜﺎل‪ :2‬ﻟﺘﻜﻥ )‪ (Un‬ﺍﻟﻤﺘﺘﺎﻟﻴﺔ ﺍﻟﻤﻌﺭﻓﺔ ﻋﻠﻰ ‪ N‬ﻜﻤﺎ ﻴﻠﻲ‪:‬‬ ‫‪U0 = 3‬‬ ‫‪+Un‬‬ ‫‪3‬‬ ‫‪Un+1 = 2‬‬ ‫* ﺍﺤﺴﺏ ‪U3،U2،U1‬‬‫ﺍﻟﻤﺘﺘﺎﻟﻴﺔ )‪ (Un‬ﻤﻌﺭﻓﺔ ﻤﻥ ﺍﻟﺸﻜل‪Un+1=f(Un) :‬‬‫ﺤﻴﺙ ‪ f‬ﻫﻲ ﺍﻟﺩﺍﻟﺔ ﺍﻟﻤﺭﻓﻘﺔ ﺒﻬﺎ ﻭﺍﻟﻤﻌﺭﻓﺔ ﻤﻥ ﺃﺠل ﻜل ‪ n‬ﻤﻭﺠﺒﺎ ﺏ‬ ‫)‪f(x‬‬ ‫=‬ ‫‪3‬‬ ‫‪+x‬‬ ‫‪2‬‬ ‫ﻭﻤﻨﻪ ﻟﺤﺴﺎﺏ ‪U3،U2،U1‬‬ ‫‪3‬‬ ‫‪Un+1 = f(Un) = 2 +Un‬‬‫= )‪U1 = f(U0‬‬ ‫‪3‬‬ ‫= ‪+ U0‬‬ ‫= ‪3 +2‬‬ ‫‪7‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫‪2‬‬‫‪U2 = f(U1) = 3 +U1 = 3 + 7 = 5‬‬ ‫‪2 22‬‬‫= )‪U3 = f(U2‬‬ ‫‪3‬‬ ‫= ‪+U2‬‬ ‫‪3‬‬ ‫‪+5‬‬ ‫=‬ ‫‪13‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫‪2‬‬

‫‪ -3‬ﺇﺘﺠﺎﻩ ﺘﻐﻴﺭ ﻤﺘﺘﺎﻟﻴﺔ \"ﺭﺘﺎﺒﺔ ﻤﺘﺘﺎﻟﻴﺔ\" ‪:‬‬ ‫ﻟﺘﻜﻥ )‪ (Un‬ﻤﺘﺘﺎﻟﻴﺔ ﻋﺩﺩﻴﺔ‬ ‫ﻨﻘﻭل ﺃﻥ‪:‬‬ ‫‪-‬ﺍﻟﻤﺘﺘﺎﻟﻴﺔ )‪ (Un‬ﻤﺘﺯﺍﻴﺩﺓ)ﺍﺒﺘﺩﺍﺀﺍ ﻤﻥ ﻤﺭﺘﺒﺔ ﻤﻌﻴﻨﺔ ‪(n0‬‬ ‫ﺇﺫﺍ ﻜﺎﻥ ‪ Un+1 ≥ Un‬ﻤﻥ ﺃﺠل ﻜل ‪n ≥ n0‬‬ ‫‪-‬ﺍﻟﻤﺘﺘﺎﻟﻴﺔ )‪ (Un‬ﻤﺘﺯﺍﻴﺩﺓ ﺘﻤﺎﻤﺎ )ﺍﺒﺘﺩﺍﺀﺍ ﻤﻥ ﻤﺭﺘﺒﺔ ﻤﻌﻴﻨﺔ ‪(n0‬‬ ‫ﺇﺫﺍ ﻜﺎﻥ ‪ Un+1 > Un‬ﻤﻥ ﺃﺠل ﻜل ‪n ≥ n0‬‬ ‫‪-‬ﺍﻟﻤﺘﺘﺎﻟﻴﺔ )‪ (Un‬ﻤﺘﻨﺎﻗﺼﺔ )ﺍﺒﺘﺩﺍﺀﺍ ﻤﻥ ﻤﺭﺘﺒﺔ ﻤﻌﻴﻨﺔ ‪(n0‬‬ ‫ﺇﺫﺍ ﻜﺎﻥ ‪ Un+1 ≤ Un‬ﻤﻥ ﺃﺠل ﻜل ‪n ≥ n0‬‬ ‫‪-‬ﺍﻟﻤﺘﺘﺎﻟﻴﺔ )‪ (Un‬ﻤﺘﻨﺎﻗﺼﺔ ﺘﻤﺎﻤﺎ)ﺍﺒﺘﺩﺍﺀﺍ ﻤﻥ ﻤﺭﺘﺒﺔ ﻤﻌﻴﻨﺔ ‪(n0‬‬ ‫ﺇﺫﺍ ﻜﺎﻥ ‪ Un+1< Un‬ﻤﻥ ﺃﺠل ﻜل ‪n ≥ n0‬‬ ‫‪ -‬ﺍﻟﻤﺘﺘﺎﻟﻴﺔ )‪ (Un‬ﺭﺘﻴﺒﺔ )ﺍﺒﺘﺩﺍﺀﺍ ﻤﻥ ﻤﺭﺘﺒﺔ ﻤﻌﻴﻨﺔ ‪(n0‬‬ ‫ﺇﺫﺍ ﻜﺎﻨﺕ ﻤﺘﻨﺎﻗﺼﺔ ﺃﻭ ﻤﺘﺯﺍﻴﺩﺓ ﻤﻥ ﺃﺠل ‪n ≥ n0‬‬‫‪ -‬ﺍﻟﻤﺘﺘﺎﻟﻴﺔ )‪ (Un‬ﺜﺎﺒﺘﺔ ﻤﻥ ﺃﺠل ‪ n‬ﻤﻥ ‪ D‬ﻤﺠﺎل ﺘﻌﺭﻴﻑ) ‪ (Un‬ﺒﺤﻴﺙ ‪ Un+1 = Un‬ﻤﻥ ﺃﺠل ﻜل ‪n ≥ n0‬‬ ‫*ﻤﻼﺤﻅﺎﺕ ‪:‬‬ ‫‪ -1‬ﻫﻨﺎﻙ ﻤﺘﺘﺎﻟﻴﺎﺕ ﻟﻴﺴﺕ ﺭﺘﻴﺒﺔ ﻤﺜل ‪Un = (-1)n‬‬‫‪ Un = -1‬ﺇﺫﺍ ﻜﺎﻥ ‪ n‬ﻓﺭﺩﻴﺎ‬ ‫ﺤﻴﺙ‬‫‪ Un = 1‬ﺇﺫﺍ ﻜﺎﻥ ‪ n‬ﺯﻭﺠﻴﺎ‬‫‪ -2‬ﺒﺈﻤﻜﺎﻨﻨﺎ ﺍﻟﻘﻭل ﺃﻥ ﺍﻟﻤﺘﺘﺎﻟﻴﺔ ﺭﺘﻴﺒﺔ ﻋﻠﻰ ﺍﻟﻤﺠﺎل [‪ [n0، ∞+‬ﺒﺩﻻ ﻤﻥ ﻗﻭﻟﻨﺎ ﺃﻨﻬﺎ ﺭﺘﻴﺒﺔ ﺍﺒﺘﺩﺍﺀﺍ ﻤﻥ‬ ‫ﻤﺭﺘﺒﺔ ﻤﻌﻴﻨﺔ ‪.n0‬‬ ‫ﺃﻤﺜﻠﺔ‪:‬‬‫ﺍﻟﻤﺜﺎل ﺍﻷﻭل‪ :‬ﺃﺩﺭﺱ ﺍﺘﺠﺎﻩ ﺘﻐﻴﺭ ﺍﻟﻤﺘﺘﺎﻟﻴﺔ )‪ (Un‬ﺍﻟﻤﻌﺭﻓﺔ ﺒﺤﺩﻫﺎ ﺍﻟﻌﺎﻡ‪:‬‬ ‫‪1‬‬ ‫) ‪ Un = E ( n‬ﺤﻴﺙ‬‫‪ n‬ﻤﻥ *‪N‬‬ ‫‪1‬‬ ‫ل‬ ‫ﺍﻟﺼﺤﻴﺢ‬ ‫ﺍﻟﺠﺯﺀ‬ ‫ﻴﻤﺜل‬ ‫(‪E‬‬ ‫‪1‬‬ ‫)‬ ‫‪ n‬ﻤﻥ ﺃﺠل‬ ‫‪n‬‬

‫ﺍﻟﺠﻭﺍﺏ‪ :‬ﻤﻥ ﺃﺠل ﻜل ‪ n ≥ 1‬ﻟﺩﻴﻨﺎ‪:‬‬ ‫‪1‬‬ ‫‪U1 = E( 1 ) = 1‬‬ ‫‪1‬‬‫‪U2 = E( 2 ) = E(0.5) = 0‬‬ ‫‪1‬‬‫‪U3 = E( 3 ) = E(0.33) = 0‬‬‫ﺍﺫﻥ ﻜﻥ ﺃﺠل ﻜل ‪ n ≥ 2‬ﻓﺈﻥ ‪:‬‬‫‪Un‬‬ ‫=‬ ‫(‪E‬‬ ‫‪1‬‬ ‫)‬ ‫=‬ ‫‪0‬‬ ‫‪n‬‬‫ﻭﻤﻨﻪ ﺍﻟﻤﺘﺘﺎﻟﻴﺔ ﻤﺘﻭﻗﻔﺔ ﻋﻠﻰ *‪N‬‬‫ﺍﻟﻤﺜﺎل‪ (Un) :2‬ﻤﺘﺘﺎﻟﻴﺔ ﻤﻌﺭﻓﺔ ﻋﻠﻰ‪:‬‬ ‫‪U0= 2‬‬ ‫‪Un+1=4Un-6‬‬‫‪ -1‬ﺍﺤﺴﺏ ‪ U3،U2،U1‬ﻤﺎﺫﺍ ﺘﻼﺤﻅ؟‬‫‪ -2‬ﻫل ﻴﻤﻜﻥ ﺃﻥ ﻨﻌﻤﻡ ﺍﻟﻤﻼﺤﻅﺔ )ﻤﻥ ﺍﻟﺴﺅﺍل‪ (1‬ﻋﻠﻰ ‪Un‬؟‬ ‫ﺍﻷﺠﻭﺒﺔ ‪:‬‬ ‫ﺤﺴﺎﺏ ‪U3، U2،U1‬‬ ‫‪U1 = 4U0 -6 = 2‬‬ ‫‪U2 = 4U1-6 = 2‬‬ ‫‪U3 = 4U2 -6 = 2‬‬‫ﻨﻼﺤﻅ ﺃﻥ ‪U1 = U2 = U3 = 2‬‬‫ﻫل ﻴﻤﻜﻥ ﺃﻥ ﻨﻌﻤﻡ ﻭﻨﻘﻭل ﺃﻥ ﻤﻥ ﺃﺠل ﻜل ‪ n‬ﻤﻥ‪ Un =2 ، N‬ﺃﻱ ﻫل )‪ (Un‬ﺜﺎﺒﺘﺔ؟‬ ‫‪-2‬‬‫ﻴﻤﻜﻥ ﺃﻥ ﻨﺒﺭﻫﻥ ﻋﻠﻰ ﺫﻟﻙ ﺒﺎﻟﺘﺭﺍﺠﻊ ﻤﻥ ﺃﺠل ﻜل ‪ n‬ﻤﻥ ‪N‬‬‫ﻨﺴﻤﻲ )‪ p(n‬ﺍﻟﺨﺎﺼﻴﺔ ﺍﻟﻤﺘﻌﻠﻘﺔ ﺒﺎﻟﻌﺩﺩ ﺍﻟﻁﺒﻴﻌﻲ ‪n‬‬‫)‪ P(n‬ﺘﻜﺎﻓﺊ ‪Un = 2 :‬‬‫‪ -1‬ﻨﺘﺤﻘﻕ ﻤﻥ ﺼﺤﺔ )‪ p(n0‬ﺤﻴﺙ ‪n0 = 0‬‬‫ﻟﺩﻴﻨﺎ ‪ U0 = 2‬ﻤﺤﻘﻘﺔ‬‫‪ -2‬ﻨﻔﺭﺽ ﺃﻥ )‪ p(n‬ﻤﻥ ﺃﺠل ﺍﻟﻤﺭﺘﺒﺔ ‪ m‬ﺤﻴﺙ ‪ m ≥ n0‬ﻭﻨﺒﺭﻫﻥ ﺼﺤﺔ )‪p(m+1‬‬‫)‪ P(m‬ﺼﺤﻴﺤﺔ ﺃﻱ ‪Um = 2‬‬

‫‪Um+1 = 4Um-6‬‬ ‫ﻟﺩﻴﻨﺎ‪:‬‬ ‫‪ Um+1= 4(2)-6‬ﺤﺴﺏ ﻓﺭﻀﻴﺔ ﺍﻟﺘﺭﺍﺠﻊ‬ ‫ﺇﺫﻥ‬ ‫ﻭﻤﻨﻪ ‪Um+1 = 8-6 = 2‬‬ ‫ﺇﺫﻥ )‪ p(m+1‬ﺼﺤﻴﺤﺔ ﻭﻤﻨﻪ ﻤﻥ ﺃﺠل ﻜل ‪ n‬ﻓﺈﻥ )‪ p(n‬ﺼﺤﻴﺤﺔ‬ ‫ﺇﺫﻥ )‪ (Un‬ﻤﺘﺘﺎﻟﻴﺔ ﺜﺎﺒﺘﺔ‬ ‫ﻭﺘﺤﻘﻕ ‪U0 = U1 = U2 =…= Un = 2‬‬‫ﻤﺜﺎل‪ : 3‬ﻟﺘﻜﻥ ﺍﻟﻤﺘﺘﺎﻟﻴﺔ )‪ (Un‬ﺍﻟﻤﻌﺭﻓﺔ ﻋﻠﻰ *‪ N‬ﺒﻌﺒﺎﺭﺓ ﺤﺩﻫﺎ ﺍﻟﻬﺎﻡ ‪:‬‬ ‫= ‪Un‬‬ ‫‪n‬‬ ‫‪n 1‬‬‫ﺍﻷﺠﻭﺒﺔ ‪ :‬ﺒﻤﺎ ﺃﻥ )‪ (Un‬ﻤﻭﺠﺒﺔ ﻤﻥ ﺍﺠل ﻜل ‪ n‬ﻤﻥ*‪ N‬ﻴﻤﻜﻨﻨﺎ ﻤﻘﺎﺭﻨﺔ‬ ‫‪.1‬‬ ‫ﺒﺎﻟﻌﺩﺩ‬ ‫‪Un  1‬‬ ‫‪Un‬‬‫‪Un  1‬‬ ‫‪n 1‬‬ ‫‪x‬‬ ‫‪n 1‬‬ ‫ﻭﻤﻨﻪ‬ ‫= ‪Un+1‬‬ ‫‪n 1‬‬ ‫= ‪Un‬‬ ‫‪n2‬‬ ‫‪n‬‬ ‫ﻟﺩﻴﻨﺎ ‪n  2‬‬ ‫‪Un  1 n2  2n  1‬‬ ‫)‪Un = n(n  2‬‬‫ﻭﻟﺩﻴﻨﺎ ‪:‬‬ ‫‪UUnn11‬‬ ‫ﺍﻟﻔﺭﻕ‬ ‫ﺇﺸﺎﺭﺓ‬ ‫ﻨﺩﺭﺱ‬ ‫ﺃﻱ‬ ‫ﺒﺎﻟﻌﺩﺩ‪.1‬‬ ‫‪Un  1‬‬ ‫ﻨﻘﺎﺭﻥ‬ ‫‪Un‬‬ ‫ ‪Un‬‬ ‫‪1‬‬ ‫‪-‬‬ ‫‪1‬‬ ‫=‬ ‫‪n2  2n 1‬‬ ‫=‪-1‬‬ ‫‪1‬‬ ‫‪Un‬‬ ‫)‪n(n  2‬‬ ‫)‪n(n  2‬‬ ‫‪Un  1‬‬ ‫ﺇﺫﻥ ‪> 1‬‬ ‫‪1‬‬ ‫‪>0‬‬ ‫ﺒﻤﺎ ﺃﻥ‬ ‫‪Un‬‬ ‫)‪n(n  2‬‬ ‫ﻭﻤﻨﻪ )‪ (Un‬ﻤﺘﺘﺎﻟﻴﺔ ﻤﺘﺯﺍﻴﺩﺓ ﺘﻤﺎﻤﺎ ﻤﻥ ﺃﺠل ‪ n‬ﻤﻥ *‪N‬‬ ‫ﻁﺭﻴﻘﺔ ﺍﻟﻌﻤل‪:‬‬‫ﺇﺫﺍ ﻜﺎﻨﺕ )‪ (Un‬ﻤﺘﺘﺎﻟﻴﺔ ﻤﻭﺠﺒﺔ ﻤﻥ ﺃﺠل ﻜل ‪) n ≥ n0‬ﻜل ﺤﺩﻭﺩﻫﺎ ﻤﻭﺠﺒﺔ( ﻟﺩﺭﺍﺴﺔ ﺍﺘﺠﺎﻩ ﺘﻐﻴﺭ ﺍﻟﻤﺘﺘﺎﻟﻴﺔ‬ ‫ﺒﺎﻟﻌﺩﺩ‪1‬‬ ‫‪Un  1‬‬ ‫)‪ (Un‬ﻴﻜﻔﻲ ﻤﻘﺎﺭﻨﺔ ﺍﻟﻌﺩﺩ‬ ‫‪Un‬‬ ‫ﻨﻘﻭل ﺃﻥ )‪ (Un‬ﻤﺘﺯﺍﻴﺩﺓ‬ ‫‪Un  1‬‬ ‫‪-1‬‬ ‫ﺇﺫﺍ ﻜﺎﻥ ‪≥ 0‬‬ ‫‪Un‬‬ ‫ﻨﻘﻭل ﺃﻥ )‪ (Un‬ﻤﺘﻨﺎﻗﺼﺔ‬ ‫‪Un  1‬‬ ‫‪-1‬‬ ‫ﺇﺫﺍ ﻜﺎﻥ ‪≤ 0‬‬ ‫‪Un‬‬

‫‪ - 4‬ﺍﻟﻤﺘﺘﺎﻟﻴﺎﺕ ﺍﻟﺤﺴﺎﺒﻴﺔ ﻭﺍﻟﻬﻨﺩﺴﻴﺔ ‪:‬‬‫ﺍﻟﺠﺩﻭل ﺍﻟﺘﺎﻟﻲ ﻴﻠﺨﺹ ﺍﻟﻘﻭﺍﻨﻴﻥ ﺍﻟﺨﺎﺼﺔ ﺒﺎﻟﻤﺘﺘﺎﻟﻴﺎﺕ ﺍﻟﺤﺴﺎﺒﻴﺔ ﻭﺍﻟﻬﻨﺩﺴﻴﺔ ﺍﻟﺘﻲ ﺩﺭﺴﺕ ﻓﺱ ﺍﻟﺴﻨﺔ ﺍﻟﺜﺎﻨﻴﺔ‬ ‫ﺜﺎﻨﻭﻱ‪.‬‬‫ﺍﻟﻤﺘﺘﺎﻟﻴﺔ ﺍﻟﻬﻨﺩﺴﻴﺔ‬ ‫ﺍﻟﻤﺘﺘﺎﻟﻴﺔ ﺍﻟﺤﺴﺎﺒﻴﺔ‬‫)‪ (Un‬ﻤﺘﺘﺎﻟﻴﺔ ﺤﺴﺎﺒﻴﺔ ﺤـﺩﻫﺎ ﺍﻷﻭل ‪ Uα‬ﻭﺃﺴﺎﺴـﻬﺎ ‪ (Vn)r‬ﻤﺘﺘﺎﻟﻴﺔ ﻫﻨﺩﺴﻴﺔ ﺤـﺩﻫﺎ ﺍﻷﻭل ‪ Vβ‬ﻭﺃﺴﺎﺴـﻬﺎ ‪q‬‬ ‫ﺤﻴﺙ‪ β‬ﻋﻨﺼﺭ ﻤﻥ ‪.N‬‬ ‫ﺤﻴﺙ‪ α‬ﻋﻨﺼﺭ ﻤﻥ ‪.N‬‬ ‫ﻋﺒﺎﺭﺓ ﺍﻟﺤﺩ ﺍﻟﻌﺎﻡ ‪ q ≠ 0‬ﻭ‪q ≠ 1‬‬ ‫ﻋﺒﺎﺭﺓ ﺍﻟﺤﺩ ﺍﻟﻌﺎﻡ ﻟﻤﺎ ‪r ≠ 0‬‬ ‫‪Vn = Vβ.qn-β‬‬ ‫‪Un = Uα+ (n-α) r‬‬ ‫ﻤﺠﻤﻭﻉ ﺤﺩﻭﺩ ﺍﻟﻤﺘﺘﺎﻟﻴﺔ ﺍﻟﻬﻨﺩﺴﻴﺔ‬ ‫ﻤﺠﻤﻭﻉ ﺤﺩﻭﺩ ﺍﻟﻤﺘﺘﺎﻟﻴﺔ ﺍﻟﺤﺴﺎﺒﻴﺔ‬ ‫‪Sn = Vβ+Vβ+1+…+Vn‬‬ ‫‪Sn = Uα+Uα+1+…+Un‬‬ ‫ﻋﺩﺩ ﺍﻟﺤﺩﻭﺩ ‪n-β+1‬‬‫‪> @Sn‬‬ ‫ﺒﻤﺎ ﺃﻥ ﻋﺩﺩ ﺍﻟﺤﺩﻭﺩ ﻫﻭ)‪(n-α+1‬‬ ‫‪Vβ‬‬ ‫)‪1q(nβ 1‬‬ ‫)‪(nα 1‬‬ ‫‪1 q‬‬ ‫‪2‬‬ ‫‪> @Sn‬‬ ‫‪Uα U n‬‬ ‫ﻓﺈﻥ‪:‬‬ ‫ﺤﺎﻟﺔ ﺨﺎﺼﺔ ‪q=1‬‬ ‫ﺤﺎﻟﺔ ﺨﺎﺼﺔ ‪r = 1‬‬ ‫‪ Vβ=Vn‬ﻓﺈﻥ )‪ (Vn‬ﺜﺎﺒﺘﺔ‬ ‫)‪ (Un‬ﺜﺎﺒﺘﺔ ‪Un = Uα‬‬ ‫‪Sn= (n-β+1)V β‬‬ ‫‪Sn = (n-α+1)Uα‬‬ ‫ﺍﻟﻭﺴﻁ ﺍﻟﺤﺴﺎﺒﻲ‬‫ﺍﻟﻬﻨﺩﺴـﻴﺔ)‪b،(Vn‬‬ ‫ﺍﻟﻤﺘﺘﺎﻟﻴﺔ‬ ‫ﻤﻥ‬ ‫ﺍﻟﻬﻨﺩﺴﻲ‬ ‫ﺍﻟﻭﺴﻁ‬ ‫ﺍﻟﺤـﺴﺎﺒﻴﺔ)‪b،(Un‬‬ ‫ﺍﻟﻤﺘﺘﺎﻟﻴﺔ‬ ‫ﺤﺩﻭﺩ ﻤﺘﺘﺎﺒﻌﺔ‬ ‫‪c،b،a‬‬ ‫‪ c،b،a‬ﺤﺩﻭﺩ ﻤﺘﺘﺎﺒﻌﺔ ﻤﻥ‬ ‫ﻫﻭ ﺍﻟﻭﺴﻁ ﺍﻟﻬﻨﺩﺴﻲ ﻭﺘﺤﻘﻕ‬ ‫ﻫﻭ ﺍﻟﻭﺴﻁ ﺍﻟﺤﺴﺎﺒﻲ ﻭﺘﺤﻘﻕ ‪:‬‬ ‫‪b2=a.c‬‬ ‫‪2b = a+c‬‬‫ﺘﺤﻘﻕ‪v v v‬‬ ‫ﺍﻟﻬﻨﺩﺴﻴﺔ‬ ‫ﺍﻟﻤﺘﺘﺎﻟﻴﺔ‬ ‫‪U1-U0‬ﺤﺩﻭﺩ‬ ‫ﺍﻟﺤﺴﺎﺒﻴﺔ ﺘﺤﻘﻕ‬ ‫ﺍﻟﻤﺘﺘﺎﻟﻴﺔ‬ ‫ﺤﺩﻭﺩ‬ ‫‪.=Un+1-Un‬‬ ‫‪=U2-U1‬‬ ‫‪=.‬‬‫= ‪v1‬‬ ‫‪v v2 =… = n1‬‬ ‫♦ﻤﺜﺎل‪ :‬ﺍﻟﻤﺘﺘﺎﻟﻴﺔ ﺍﻟﺤﺴﺎﺒﻴﺔ )‪ (Un‬ﻤﺘﺘﺎﻟﻴﺔ ﺤﺴﺎﺒﻴﺔ ﺤﺩﻫﺎ‬ ‫‪0‬‬ ‫‪1n‬‬ ‫ﺍﻷﻭل ‪ U0‬ﻭﺃﺴﺎﺴﻬﺎ ‪r‬‬ ‫‪-1‬ﻋﻴﻥ ﺃﺴﺎﺱ ﺍﻟﻤﺘﺘﺎﻟﻴﺔ )‪(Un‬‬‫‪-2‬ﺃﺤﺴﺏ ‪ U0‬ﺜﻡ ﺃﻋﻁ ﻋﺒﺎﺭﺓ ‪ Un‬ﺒﺩﻻﻟﺔ ‪ n‬ﻋﻠﻤـﺎ ﺃﻥﺇﺩﺍﻜﺎﻥ ‪ q=0‬ﻓﺈﻥ )‪ (Vn‬ﻤﺘﺘﺎﻟﻴﺔ ﻤﺘﻭﻗﻔﺔ‬ ‫‪U13 = 22‬‬ ‫‪U7 = 10‬‬

‫♦ﻤﺜﺎل‪ :‬ﺍﻟﻤﺘﺘﺎﻟﻴﺎﺕ ﺍﻟﻬﻨﺩﺴﻴﺔ )‪ (Vn‬ﻤﺘﺘﺎﻟﻴـﺔ ﻫﻨﺩﺴـﻴﺔ‬ ‫‪-3‬ﺍﺤﺴﺏ ﺍﻟﻤﺠﻤﻭﻉ ‪ S‬ﺤﻴﺙ‬ ‫‪S=U0+U1+…+U99‬‬‫ﺠﻤﻴﻊ ﺤﺩﻭﺩﻫﺎ ﻤﻭﺠﺒﺔ ﺤﻴﺙ *‪n є N‬‬ ‫‪x‬ﺍﻷﺠﻭﺒﺔ‬ ‫‪U13 = 22 ، U7=10‬‬ ‫ﺒﺤﻴﺙ ‪V4 x V6=16‬‬ ‫ﻟﺩﻴﻨﺎ‬‫‪1‬‬ ‫‪U13 = U7+(13-7)r‬‬‫‪ -1‬ﺃﺤﺴﺏ ﺍﻷﺴﺎﺱ ‪ q‬ﻋﻠﻤﺎ ﺃﻥ =‪4 V1‬‬ ‫‪22=10+6r‬‬‫‪-2‬ﺍﺤﺴﺏ ﻋﺒﺎﺭﺓ ‪ Vn‬ﺒﺩﻻﻟﺔ ‪n‬‬ ‫‪-3‬ﺃﺤﺴﺏ ‪ Sn‬ﺤﻴﺙ‬ ‫‪Sn=V1+V2+…+Vn‬‬ ‫• ﺍﻷﺠﻭﺒﺔ ‪:‬‬ ‫‪22-10=6r‬‬ ‫ﻭﻤﻨﻪ‬ ‫ﺍﺫﻥ ‪:‬‬ ‫ﻟﺩﻴﻨﺎ ‪V4 x V6 = 16‬‬ ‫‪r‬‬ ‫‪12‬‬ ‫‪2‬‬‫‪ V5x‬ﻫﻭ ﺍﻟﻭﺴﻁ ﺍﻟﻬﻨﺩﺴﻲ ﻟﻠﺤﺩﻴﻥ ‪ V6‬ﻭ‪ V5‬ﻭﻤﻨﻪ ‪V52‬‬ ‫‪6‬‬ ‫‪ = V4 x V6‬ﺃﻱ ‪V52 = 16‬‬ ‫ﺇﺫﻥ ﺍﻷﺴﺎﺱ ‪r =2‬‬‫ﻭﻤﻨﻪ ‪ V5 = 4‬ﻷﻥ ﺍﻟﺤﺩﻭﺩ ﻤﻭﺠﺒﺔ‬ ‫‪x‬ﺤﺴﺎﺏ ‪U0‬‬‫‪U1‬‬ ‫‪1‬‬ ‫ﺤﺴﺎﺏ ‪ q‬ﻋﻠﻤﺎ ﺃﻥ‬ ‫‪x‬‬ ‫ﻟﺩﻴﻨﺎ ‪:‬‬ ‫‪4‬‬ ‫‪U7= U0+7r‬‬ ‫‪U5 = U1.q4‬‬ ‫ﻟﺩﻴﻨﺎ‬ ‫)‪ 10 = U0+7(2‬ﻭﻤﻨﻪ‬ ‫‪U0 = 10-14 = -4‬‬ ‫‪4 = 1 .q4‬‬ ‫ﻭﻤﻨﻪ‬ ‫‪4‬‬ ‫‪x‬ﻋﺒﺎﺭﺓ ‪ Un‬ﺒﺩﻻﻟﺔ ‪n‬‬‫‪q4 = 4 x 4= 16‬‬ ‫ﺃﻱ‬ ‫‪Un = U0+nr‬‬ ‫ﻟﺩﻴﻨﺎ‬‫‪q4 = 24‬‬ ‫ﻭﻤﻨﻪ ‪Un = -4+2n :‬‬ ‫‪q=2‬‬ ‫ﻭﻤﻨﻪ‬ ‫‪x‬ﺤﺴﺎﺏ ﺍﻟﻤﺠﻤﻭﻉ‪:‬‬ ‫‪S = U0+U1+…+U99‬‬ ‫‪ x‬ﻋﺒﺎﺭﺓ ‪ Vn‬ﺒﺩﻻﻟﺔ ‪n‬‬ ‫ﻟﺩﻴﻨﺎ ‪Vn = V1.qn-1‬‬ ‫ﻋﺩﺩ ﺍﻟﺤﺩﻭﺩ ‪99-0+1 = 100‬‬‫= ‪Vn‬‬ ‫‪1‬‬ ‫‪(2)n-1‬‬ ‫ﻭﻤﻨﻪ‬ ‫‪100‬‬ ‫‪4‬‬ ‫= ‪[U0+U99] 2 S‬‬‫ﺤﺴﺎﺏ ‪> @Sn= v1+v2+..+vn‬‬ ‫‪x‬‬ ‫])‪Sn = 50[-4+(-4)+2(99‬‬ ‫‪x‬‬ ‫‪Sn‬‬ ‫‪V1‬‬ ‫‪1q n‬‬ ‫ﻭﻤﻨﻪ ‪S = 9500‬‬ ‫‪1 q‬‬ ‫][‬ ‫‪1 2n‬‬ ‫‪1‬‬ ‫‪x‬‬ ‫‪1 2‬‬ ‫= ‪4 Sn‬‬ ‫‪Sn = -‬‬ ‫‪1‬‬ ‫]‪[1-2n‬‬ ‫‪4‬‬

‫‪ -5‬ﺇﺘﺠﺎﻩ ﺘﻐﻴﺭ ﻤﺘﺘﺎﻟﻴﺔ ﺤﺴﺎﺒﻴﺔ ﻭﻤﺘﺘﺎﻟﻴﺔ ﻫﻨﺩﺴﻴﺔ‬‫أ( ﺇﺘﺠﺎﻩ ﺘﻐﻴﺭ ﻤﺘﺘﺎﻟﻴﺔ ﺤﺴﺎﺒﻴﺔ ‪:‬‬‫)‪ (Un‬ﻤﺘﺘﺎﻟﻴﺔ ﺤﺴﺎﺒﻴﺔ ﻤﻌﺭﻓﺔ ﻤﻊ ‪ N‬ﺤﺩﻫﺎ ﺍﻷﻭل ‪ U0‬ﻭﺃﺴﺎﺴﻬﺎ ‪.r‬‬‫ﻟﺩﻴﻨﺎ ‪ Un = U0 + nr‬ﻭﻤﻨﻪ ‪Un+1 = U0 +(n+1)r‬‬‫ﻨﺩﺭﺱ ﺇﺸﺎﺭﺓ ‪( Un+1 – Un) :‬‬‫]‪Un+1 – Un = [U0 +(n+1)r] - [U0 + nr‬‬ ‫ﻟﺩﻴﻨﺎ‪:‬‬‫‪= U0 + nr + r-U0 - nr‬‬‫‪Un+1 – Un = r‬‬ ‫ﺇﺫﻥ‬ ‫ﺇﺫﻥ ﺇﺘﺠﺎﻩ ﺍﻟﻤﺘﺘﺎﻟﻴﺔ ﺍﻟﺤﺴﺎﺒﻴﺔ ﻴﻌﺘﻤﺩ ﻋﻠﻰ ﺇﺸﺎﺭﺓ ﺃﺴﺎﺴﻬﺎ ‪r‬‬‫‪ r > 0 x‬ﻴﻜﺎﻓﺊ )‪ (Un‬ﻤﺘﺯﺍﻴﺩﺓ ﺘﻤﺎﻤﺎ ﻤﻥ ﺃﺠل ﻜل ‪n є N‬‬‫‪ r < 0 x‬ﻴﻜﺎﻓﺊ )‪ (Un‬ﻤﺘﻨﺎﻗﺼﺔ ﺘﻤﺎﻤﺎ ﻤﻥ ﺃﺠل ﻜل ‪n є N‬‬‫‪ r = 0 x‬ﻴﻜﺎﻓﺊ )‪ (Un‬ﺜﺎﺒﺘﺔ ﻤﻥ ﺃﺠل ﻜل ‪ n‬ﻤﻥ ‪N‬‬ ‫ﻤﺜﺎل ‪ :‬ﻟﺘﻜﻥ )‪ (Un‬ﺍﻟﻤﺘﺘﺎﻟﻴﺔ ﺍﻟﻤﻌﺭﻓﺔ ﺒﺤﺩﻫﺎ ﺍﻷﻭل ‪U0 = -4‬‬ ‫ﻭﺒﺎﻟﻌﻼﻗﺔ ‪ Un+1 = Un-2n + 3‬ﻤﻥ ﺃﺠل ﻜل ﻋﺩﺩ ﻁﺒﻴﻌﻲ ‪n‬‬ ‫ﻭﻟﺘﻜﻥ )‪ (rn‬ﺍﻟﻤﺘﺘﺎﻟﻴﺔ ﺍﻟﻤﻌﺭﻓﺔ ﻤﻥ ﺃﺠل ﻜل ﻋﺩﺩ ﻁﺒﻴﻌﻲ ‪ n‬ﺒﺎﻟﻌﻼﻗﺔ‬ ‫‪rn = Un+1 – Un‬‬‫‪ -1‬ﺃﺜﺒﺕ ﺃﻥ )‪ (rn‬ﻤﺘﺘﺎﻟﻴﺔ ﺤﺴﺎﺒﻴﺔ ﻴﻁﻠﺏ ﺘﻌﻴﻴﻥ ﺃﺴﺎﺴﻬﺎ ﻭﺤﺩﻫﺎ ﺍﻷﻭل‪.‬‬‫‪ -2‬ﺍﺴﺘﻨﺘﺞ ﺍﺘﺠﺎﻩ ﺘﻐﻴﺭ )‪(rn‬‬

‫ﺍﻷﺠﻭﺒﺔ‬ ‫)‪ (rn‬ﻤﺘﺘﺎﻟﻴﺔ ﺤﺴﺎﺒﻴﺔ ﺃﺴﺎﺴﻬﺎ ‪ r‬ﻴﻜﺎﻓﺊ ﻤﻥ ﺃﺠل ﻜل ﻋﺩﺩ ‪ n‬ﻤﻥ ‪N‬‬ ‫‪rn+1 – rn = r‬‬ ‫ﻨﺤﺴﺏ ‪Un+1 – Un :‬‬ ‫ﻟﺩﻴﻨﺎ‪:‬‬ ‫‪Un+1-Un = Un – 2n + 3 – Un‬‬ ‫‪= -2n + 3‬‬ ‫ﻭﻤﻨﻪ ﻨﺤﺴﺏ ‪rn+1 – rn‬‬ ‫)‪rn+1 – rn = -2(n+1) + 3 – (-2n + 3‬‬ ‫‪= -2n -2 + 3 + 2n – 3‬‬ ‫‪rn+1 – rn = -2 = r‬‬ ‫ﺇﺫﻥ )‪ (rn‬ﻤﺘﺘﺎﻟﻴﺔ ﺤﺴﺎﺒﻴﺔ ﺃﺴﺎﺴﻬﺎ ‪r = -2‬‬ ‫ﺤﺴﺎﺏ ‪: r0‬‬ ‫‪r0 = -1 + 4‬‬ ‫‪ r0 = U1 – U0‬ﻭﻤﻨﻪ‬ ‫ﺇﺫﻥ ‪r0 = 3 :‬‬ ‫ﻭﺒﻤﺎ ﺃﻥ ‪ r = -2 < 0‬ﺇﺫﻥ )‪ (rn‬ﻤﺘﺘﺎﻟﻴﺔ ﺤﺴﺎﺒﻴﺔ ﻤﺘﻨﺎﻗﺼﺔ ﺘﻤﺎﻤﺎ‪.‬‬ ‫ﺏ( ﺇﺘﺠﺎﻩ ﺘﻐﻴﺭ ﻤﺘﺘﺎﻟﻴﺔ ﻫﻨﺩﺴﻴﺔ ‪:‬‬ ‫)‪ (Un‬ﻤﺘﺘﺎﻟﻴﺔ ﻫﻨﺩﺴﻴﺔ ﺃﺴﺎﺴﻬﺎ ‪ q‬ﺤﻴﺙ ‪ q ≠ 1‬ﻭﺤﺩﻫﺎ ﺍﻷﻭل ‪.U0‬‬ ‫ﻟﺩﻴﻨﺎ ‪ Un = U0 . qn‬ﻭ ‪Un+1 = U0 qn+1‬‬ ‫ﻭﻤﻨﻪ ﻨﺤﺴﺏ ‪Un+1 – Un‬‬ ‫]‪Un+1 – Un = U0 qn [q -1‬‬ ‫ﻨﻤﻴﺯ ﺍﻟﺤﺎﻻﺕ ﺤﺴﺏ ﺍﻹﺸﺎﺭﺓ ) ‪(q-1‬‬‫‪q -∞ 0‬‬ ‫∞‪1 +‬‬‫‪q -1‬‬ ‫‪-‬‬ ‫‪-+‬‬ ‫‪ q -1 > 0 (1‬ﺃﻱ ‪q > 1‬‬‫ﺇﺫﻥ ‪ qn‬ﻤﻭﺠﺒﺎ ﻭﻤﻨﻪ ﺍﺘﺠﺎﻩ ﺘﻐﻴﺭ ﺍﻟﻤﺘﺘﺎﻟﻴﺔ ﺍﻟﻬﻨﺩﺴﻴﺔ ﻴﻌﺘﻤﺩ ﺇﻟﻰ ﺇﺸﺎﺭﺓ ‪U0‬‬‫ﺃ( ‪ q > 1‬ﻭ ‪ U0 > 0‬ﻴﻜﺎﻓﺊ )‪(Un‬ﺘﺯﺍﻴﺩﺓ ﺘﻤﺎﻤﺎ ﻤﻥ ﺃﺠل ﻜل ‪n є N‬‬‫ﺏ(‪ q < 1‬ﻭ ‪ U0 < 0‬ﻴﻜﺎﻓﺊ)‪( Un‬ﻨﺎﻗﺼﺔ ﺘﻤﺎﻤﺎ ﻤﻥ ﺃﺠل ﻜل‪n є N‬‬ ‫‪ 0< q < 1 (2‬ﺇﻥ ‪ q -1 < 0‬ﻭ ‪ qn‬ﻤﻭﺠﺒﺎ‬

‫ﻭﻤﻨﻪ ‪:‬‬ ‫ﺃ( ‪ 0<q<1‬ﻭ‪ U0 > 0‬ﻴﻜﺎﻓﺊ )‪ (Un‬ﻤﺘﻨﺎﻗﺼﺔ ﺘﻤﺎﻤﺎ ﻤﻥ ﺃﺠل ﻜل‪n є N‬‬ ‫ﺏ(‪ 0<q<1‬ﻭ ‪ U0 < 0‬ﻴﻜﺎﻓﺊ )‪ (Un‬ﻤﺘﺯﺍﻴﺩﺓ ﺘﻤﺎﻤﺎ ﻤﻥ ﺃﺠل ﻜل‪n є N‬‬‫‪ q ≤ 0 (3‬ﺇﺫﻥ ‪ q -1 < 0‬ﻭ ‪ qn‬ﻟﻴﺴﺕ ﻟﻪ ﺇﺸﺎﺭﺓ ﺜﺎﺒﺘﺔ ﻭﺫﻟﻙ ﺤﺴﺏ ﻜﻭﻥ ‪ n‬ﺯﻭﺠﻴﺎ ﺃﻭ ﻓﺭﺩﻴﺎ‪.‬‬ ‫ﻭﻤﻨﻪ )‪ (Un‬ﻟﻴﺴﺕ ﺭﺘﻴﺒﺔ‪.‬‬‫ﻤﺜﺎل ‪ (Un) :‬ﻤﺘﺘﺎﻟﻴﺔ ﻋﺩﺩﻴﺔ ﻤﻌﺭﻓﺔ ﺒﺤﺩﻫﺎ ﺍﻟﻌﺎﻡ‬‫= ‪ Un‬ﻤﻥ ﺃﺠل ﻜل ‪ n‬ﻤﻥ ‪N‬‬ ‫‪4 n 1‬‬ ‫‪7n‬‬‫‪ .1‬ﺒﺭﻫﻥ ﺃﻥ )‪ (Un‬ﻤﺘﺘﺎﻟﻴﺔ ﻫﻨﺩﺴﻴﺔ ﻋﻴﻥ ﺃﺴﺎﺴﻬﺎ ﻭﺤﺩﻫﺎ ﺍﻷﻭل ‪.U0‬‬‫‪ .2‬ﻤﺎ ﻫﻭ ﺍﺘﺠﺎﻩ ﺘﻐﻴﺭ ﺍﻟﻤﺘﺘﺎﻟﻴﺔ )‪ (Un‬؟‬ ‫ﺍﻷﺠﻭﺒﺔ ‪:‬‬ ‫‪ .1‬ﻟﺩﻴﻨﺎ ‪:‬‬‫= ‪Un‬‬ ‫‪4 n 1‬‬ ‫‪= 4( 4 )n‬‬ ‫‪7n‬‬ ‫‪7‬‬‫)‪ (Un‬ﻟﻬﺎ ﺍﻟﺸﻜل ‪ a.bn‬ﻭﻫﻭ ﻋﺒﺎﺭﺓ ﻋﻥ ﺍﻟﺤﺩ ﺍﻟﻌﺎﻡ ﻟﻤﺘﺘﺎﻟﻴﺔ ﻫﻨﺩﺴﻴﺔ ﺤﺩﻫﺎ ﺍﻷﻭل ‪ a‬ﻭﺃﺴﺎﺴﻬﺎ ‪b‬‬ ‫‪4‬‬ ‫ﻭﻤﻨﻪ ﻭﺒﺎﻟﻤﻭﺍﻓﻘﺔ ﻤﻊ ﻋﺒﺎﺭﺓ ‪ Un‬ﻨﺤﻭ ‪ U0 = 4‬ﻭ ‪q = 7‬‬ ‫‪ .2‬ﺍﺘﺠﺎﻩ ﺘﻐﻴﺭ )‪(Un‬‬ ‫ﺇﺫﻥ ‪0<q<1‬‬ ‫=‪q‬‬ ‫‪4‬‬ ‫ﻟﺩﻴﻨﺎ‬ ‫‪7‬‬‫ﻜﺫﻟﻙ ‪ U0 > 4‬ﻭﻤﻨﻪ )‪ (Un‬ﻤﺘﺘﺎﻟﻴﺔ ﻫﻨﺩﺴﻴﺔ ﻤﺘﻨﺎﻗﺼﺔ ﺘﻤﺎﻤﺎ ﻤﻥ ﺃﺠل ﻜل ‪ n‬ﻤﻥ ‪N‬‬

‫‪Un+1 – Un‬‬ ‫*ﻤﻼﺠﻅﺔ ‪ :‬ﻴﻤﻜﻥ ﺃﻥ ﻨﺩﺭﺱ ﺇﺘﺠﺎﻩ ﺘﻐﻴﺭ ﺍﻟﻤﺘﺘﺎﻟﻴﺔ )‪ (Un‬ﺒﺩﺭﺍﺴﺔ ﺇﺸﺎﺭﺓ ﺍﻟﻔﺭﻕ‬ ‫‪ Un+1 – Un = 4‬‬ ‫‪4‬‬ ‫‪n1 – 4 ( 4 )n‬‬ ‫ﻟﺩﻴﻨﺎ ‪:‬‬ ‫‪7‬‬ ‫‪7‬‬ ‫=@ >  ‬‫‪4‬‬‫‪4‬‬ ‫‪n‬‬ ‫‪741‬‬ ‫‪7‬‬ ‫‪  > @= 4‬‬‫‪4‬‬‫‪n‬‬‫‪3‬‬ ‫‪7‬‬ ‫‪7‬‬ ‫ﺇﺫﻥ ‪ Un+1– Un < 0‬ﻭﻤﻨﻪ )‪ (Un‬ﻤﺘﻨﺎﻗﺼﺔ ﺘﻤﺎﻤﺎ ﻤﻥ ﺃﺠل ﻜل ‪ n‬ﻤﻥ ‪N‬‬‫‪ -6‬ﺇﺴــﺘﻌﻤﺎل ﺍﻟﻤﺘﺘﺎﻟﻴــﺎﺕ ﺍﻟﺤــﺴﺎﺒﻴﺔ ﻭﺍﻟﻬﻨﺩﺴــﻴﺔ ﻟﺤــل ﻤــﺸﻜﻼﺕ‬ ‫ﻤﻥ ﺍﻟﺤﻴﺎﺓ ﺍﻟﻴﻭﻤﻴﺔ‬ ‫♦ﺍﻟﻤﺸﻜﻠﺔ ﺭﻗﻡ ‪: 1‬‬ ‫ﺍﻟﻤﻌﻁﻴﺎﺕ ﻭﺍﻷﺴﺌﻠﺔ ‪:‬‬‫ﻭﻀﻊ ﺘﻠﻤﻴﺫ ﻤﺒﻠﻐﺎ ﻤﻘﺩﺍﺭﻩ ‪ 6000‬ﺩ‪.‬ﺝ ﻓﻲ ﺍﻟﺒﻨﻙ ﺒﻔﻭﺍﺌﺩ ﺒﺴﻴﻁﺔ ﻟﻌﺩﺓ ﺴﻨﻭﺍﺕ‪ ،‬ﺃﻱ ﺃﻨﻪ ﻋﻨﺩ ﻨﻬﺎﻴﺔ ﻜل‬‫ﺴﻨﺔ ﻴﻤﻨﺢ ﺍﻟﺒﻨﻙ ﻓﺎﺌﺩﺓ ﻗﺩﺭﻫﺎ ‪ %8‬ﻟﻴﺯﻴﺩ ﺇﺩﺨﺎﺭﻩ ﻜل ﺴﻨﺔ ﺒﻤﺒﻠﻎ ﺜﺎﺒﺕ ﻴﺴﺎﻭﻱ ‪ %8‬ﻤﻥ ﺍﻟﻤﺒﻠﻎ ﺍﻻﺒﺘﺩﺍﺌﻲ‪.‬‬ ‫‪ -‬ﻴﺭﻴﺩ ﺍﻟﺘﻠﻤﻴﺫ ﻤﻌﺭﻓﺔ ﺍﻟﻤﺒﻠﻎ ﻟﻪ ﻜل ﺴﻨﺔ‬ ‫‪ .1‬ﺃﺤﺴﺏ ‪U3 . U2 . U1‬‬ ‫‪ .2‬ﺘﺤﻘﻕ ﺃﻨﻪ ﻤﻥ ﺃﺠل ﻜل ﻋﺩﺩ ﻁﺒﻴﻌﻲ ‪ n‬ﻟﺩﻴﻨﺎ ‪Un+1 = Un + 480‬‬ ‫‪ .3‬ﻋّﺒﺭ ﻫﻥ ‪ Un‬ﺒﺩﻻﻟﺔ ‪n‬‬‫‪ .4‬ﻤﺎ ﻫﻭ ﻋﺩﺩ ﺍﻟﺴﻨﻭﺍﺕ ﺍﻟﺘﻲ ﻴﺠﺏ ﺍﻨﺘﻅﺎﺭﻫﺎ ﻟﻴﻀ ّﻌﻑ ﺍﻟﺘﻠﻤﻴﺫ ﺍﻟﻤﺒﻠﻎ ﺍﻻﺒﺘﺩﺍﺌﻲ ﺇﻟﻰ ‪ 3‬ﻤﺭﺍﺕ ؟‬ ‫• ﺍﻷﺠﻭﺒﺔ ‪:‬‬ ‫ﻨﻀﻊ ‪ U0 = 6000‬ﺍﻟﻤﺒﻠﻎ ﺍﻻﺒﺘﺩﺍﺌﻲ‬ ‫‪6000‬‬ ‫ﺇﺫﻥ ‪100 U1 = 6000 + 8 x :‬‬ ‫‪U1 = 6000 + 480‬‬ ‫‪U1 = 6480‬‬ ‫‪U2‬‬ ‫=‬ ‫‪6480‬‬ ‫‪+‬‬ ‫‪6000‬‬ ‫‪x‬‬ ‫‪8‬‬ ‫‪100‬‬ ‫‪= 6480 + 480‬‬

‫‪U2 = 6960‬‬ ‫‪8‬‬ ‫‪U3 = 6960 + 6000 x 100‬‬ ‫‪= 6960 + 480‬‬ ‫‪U3 = 7440‬‬ ‫‪ -2‬ﺍﻟﺘﺤﻘﻕ ﺃﻨﻪ ﻤﻥ ﺃﺠل ﻜل ﻋﺩﺩ ﻁﺒﻴﻌﻲ ‪ n‬ﻟﺩﻴﻨﺎ ‪:‬‬ ‫‪Un+1 = Un + 480‬‬ ‫ﻟﺩﻴﻨﺎ‬ ‫‪U1 = U0 + 480‬‬ ‫‪U2 = U1 + 480‬‬ ‫‪U3 = U2 + 480‬‬ ‫ﻭﺒﺘﺨﻤﻴﻥ ﻨﺠﺩ ‪Un+1 = Un + 480‬‬ ‫ﻭﻤﻨﻪ )‪ (Un‬ﻤﺘﺘﺎﻟﻴﺔ ﺤﺴﺎﺒﻴﺔ ﺤﺩﻫﺎ ﺍﻷﻭل ‪ U0 = 6000‬ﻭ ﺃﺴﺎﺴﻬﺎ‬ ‫‪r = 480‬‬ ‫ﺇﺫﻥ ‪ Un = U0 + nr‬ﺃﻱ ‪Un = 6000 + 480n‬‬ ‫ﻋﺩﺩ ﺍﻟﺴﻨﻭﺍﺕ ﻟﻜﻲ ﻴﺘﻀﺎﻋﻑ ﺍﻟﻤﺒﻠﻎ ﺍﻻﺒﺘﺩﺍﺌﻲ ‪ 3‬ﻤﺭﺍﺕ‬ ‫ﻨﻀﻊ ‪Un = 3 x 6000‬‬ ‫‪6000 + 480n = 18000‬‬ ‫‪480n = 18000 – 6000‬‬ ‫‪480n = 1200‬‬ ‫ﻭﻤﻨﻪ ‪n = 25‬‬ ‫ﺇﺫﻥ ﻋﺩﺩ ﺍﻟﺴﻨﻭﺍﺕ ﻫﻭ ‪ 25‬ﺴﻨﺔ‬ ‫♦ﺍﻟﻤﺸﻜﻠﺔ ﺍﻟﺜﺎﻨﻴﺔ ‪:‬‬ ‫ﺍﻟﻤﻌﻁﻴﺎﺕ ﻭﺍﻷﺴﺌﻠﺔ ‪:‬‬‫ﻓﻲ ﺴﻨﺔ ‪ 2000‬ﻜﺎﻥ ﺴﻌﺭ ﺍﻟﻐﺭﺍﻡ ﺍﻟﻭﺍﺤﺩ ﻤﻥ ﺍﻟﺫﻫﺏ ﺍﻟﺨﺎﻟﺹ ﻴﻘ ّﺩﺭ ﺒﻘﻴﻤﺔ ‪ 1000‬ﺩ‪.‬ﺝ‪ ،‬ﻋﻠﻤﺎ ﺃﻥ‬ ‫ﺴﻌﺭ ﻫﺫﺍ ﺍﻷﺨﻴﺭ ﻴﺯﺩﺍﺩ ﻜل ﺴﻨﺔ ﺒﻤﻘﺩﺍﺭ ‪ %20‬ﻤﻥ ﺍﻟﻤﺒﻠﻎ ﺍﻟﺫﻱ ﻜﺎﻥ ﻋﻠﻴﻪ ﻓﻲ ﺍﻟﺴﻨﺔ ﺍﻟﻔﺎﺭﻁﺔ‪.‬‬‫‪ -1‬ﺍﺸﺘﺭﺕ ﺘﻠﻤﻴﺫﺓ ﺨﺎﺘﻤﺎ ﻭﺯﻨﻪ ‪4‬ﻏﺭﺍﻤﺎﺕ ﻓﻲ ﻴﻭﻡ ‪ 2000/01/01‬ﻜﻡ ﺴﻴﺒﻠﻎ ﺜﻤﻥ ﻫـﺫﺍ ﺍﻟﺨـﺎﺘﻡ ﻴـﻭﻡ‬ ‫‪ 2007/01/01‬؟‬‫‪ -2‬ﺃﺭﺍﺩﺕ ﻫﺫﻩ ﺍﻟﺘﻠﻤﻴﺫﺓ ﺃﻥ ﺘﺒﻴﻊ ﺨﺎﺘﻤﻬﺎ ﻓﻲ ﻋﺎﻡ ‪ 2007‬ﻟﺼﺎﺌﻎ ﻤﺎ ﻫﻭ ﺜﻤﻥ ﺒﻴﻥ ﻫـﺫﺍ ﺍﻟﺨـﺎﺘﻡ ﻋﻠﻤـﺎ ﺃﻥ‬ ‫ﺍﻟﺼﺎﺌﻎ ﻴﺄﺨﺫ ﻨﺴﺒﺔ ﻓﻲ ﺍﻟﺭﺒﺢ ﻤﻘﺩﺭﺓ ﺒـ‪ %20‬ﻤﻥ ﺍﻟﻤﺒﻠﻎ ﺍﻹﺠﻤﺎﻟﻲ ﻟﻠﺨﺎﺘﻡ ؟‬ ‫•ﺍﻷﺠﻭﺒﺔ ‪:‬‬ ‫ﺜﻤﻥ ﺍﻟﺨﺎﺘﻡ ﻴﻭﻡ ‪ 2000/01/01‬ﻫﻭ ‪ 4000‬ﺩ‪.‬ﺝ‬ ‫ﻨﻀﻊ ‪U0 = 4000‬‬

‫ﺍﻟﺯﻴﺎﺩﺓ ﻜل ﻋﺎﻡ ﻤﻘﺩﺭﺓ ﺒـ ‪ %20‬ﻤﻥ ﺍﻟﻤﺒﻠﻎ ﺍﻹﺠﻤﺎﻟﻲ‬‫ﺇﺫﻥ ﻴﻭﻡ ‪ 2001/01/01‬ﻴﺼﺒﺢ ﺜﻤﻥ ﺍﻟﺨﺎﺘﻡ ‪:‬‬ ‫‪20‬‬ ‫‪U1 = U0 + 4000 x 100‬‬‫‪U1 = 4800‬‬ ‫‪U1 = U0 + U0 x 0.2‬‬ ‫)‪U1 = U0 (1+0.2‬‬ ‫‪U1 = 1.02 x U0‬‬ ‫‪U1 = 1.02 x 4000‬‬ ‫ﻓﻲ ﺴﻨﺔ ‪ 2000‬ﻴﺼﺒﺢ ﺜﻤﻥ ﺍﻟﺨﺎﺘﻡ ‪:‬‬‫‪U2 = 5760‬‬ ‫‪U2 = U1 + U1 x 0.2‬‬ ‫)‪U2 = U1 (1+0.2‬‬ ‫)‪U2 = U1 (1.2‬‬ ‫)‪U2 = (U0 x 1.2)(1.2‬‬ ‫‪U2 = U0 (1.2)2‬‬ ‫)‪U2 = 4000 x (1.44‬‬‫ﻭﻫﻜﺫﺍ ﻓﺈﻥ ﺜﻤﻥ ﺍﻟﺨﺎﺘﻡ ﻴﺯﺩﺍﺩ ﺒﺄﺴﺎﺱ ﻤﺘﺘﺎﻟﻴﺔ ﻫﻨﺩﺴﻴﺔ ‪ q=1.2‬ﻭﻤﻨﻪ ‪U7 = U0 x (1.2)7‬‬ ‫‪U7 = 4000 x (1.2)7‬‬ ‫‪U7 = 14332.7‬‬ ‫ﻭﻤﻨﻪ ﺜﻤﻥ ﺒﻴﻊ ﺍﻟﺨﺎﺘﻡ ﻋﺎﻡ ‪ 2007‬ﻫﻭ ‪ 14332.7‬ﺩ‪.‬ﺝ‪.‬‬‫* ﻨﺴﺒﺔ ﺭﺒﺢ ﺍﻟﺼﺎﺌﻎ ﻤﻥ ﺍﻟﺭﺒﺢ ﺍﻹﺠﻤﺎﻟﻲ ﻫﻲ ‪ 20%‬ﺇﺫﻥ ﻫﺫﻩ ﺍﻟﻨﺴﺒﺔ ﺘﻘﺩﺭ ﺒـ‪:‬‬ ‫‪14332.7‬‬ ‫‪x‬‬ ‫‪20‬‬ ‫‪= 2866.54‬‬ ‫‪100‬‬ ‫ﻭﻤﻨﻪ ﺜﻤﻥ ﺒﻴﻊ ﺍﻟﺨﺎﺘﻡ ﻫﻭ ‪:‬‬ ‫‪14332.7 – 2866.54 = 11466.16‬‬‫ﺜﻤﻥ ﺒﻴﻊ ﺍﻟﺨﺎﺘﻡ ﻋﺎﻡ ‪ 2007‬ﻫﻭ ‪ 11466.16‬ﺩ‪.‬ﺝ‬

‫‪ -7‬ﺍﻟﻤﺘﺘﺎﻟﻴﺎﺕ ﻤﻥ ﺍﻟﺸﻜل ‪ U n1 aU n  b :‬ﻤﻊ ‪ a≠0‬ﻭ ‪b≠0‬‬ ‫أ‪ -‬ﺣﺴﺎب اﻟﺤﺪ اﻟﻌﺎم ‪ Un‬ﺒﺩﻻﻟﺔ ‪ n‬ﺤﻴﺙ ‪ n‬ﻋﺩﺩﺍ ﻁﺒﻴﻌﻴﺎ‬ ‫‪ x‬ﻨﻤﻴﺯ ﺍﻟﺤﺎﻟﺔ ﺍﻟﺨﺎﺼﺔ ‪a=1‬‬‫ﺇﺫﻥ ‪ U n1 U n b‬ﻭ ﻤﻨﻪ ‪ U n1  U n b‬ﻭ ﻤﻨﻪ )‪ (Un‬ﻤﺘﺘﺎﻟﻴﺔ ﺤﺴﺎﺒﻴﺔ ﺤﺩﻫﺎ ﺍﻷﻭل‬ ‫‪ U0‬ﻭ ﺃﺴﺎﺴﻬﺎ ‪ b‬ﻭ ﻤﻨﻪ ‪U n U 0  bn‬‬ ‫‪ x‬ﻨﻔﺭﺽ ‪a≠1‬‬‫‪ U n1  aU n‬ﻭ‬ ‫ﺇﺫﺍ ﻜﺎﻨﺕ )‪ (Un‬ﻭ )‪ (V1‬ﻤﺘﺘﺎﻟﻴﺘﺎﻥ ﺘﺤﻘﻘﺎﻥ ﺍﻟﻌﻼﻗﺔ ﺍﻟﺘﺭﺍﺠﻌﻴﺔ ‪b‬‬ ‫‪x‬‬ ‫‪Vn1  aVn b‬‬‫ﺇﺫﻥ ﻓﺭﻗﻬﺎ ‪ Wn‬ﻫﻭ ﺤﺩ ﻋﺎﻡ ﻟﻤﺘﺘﺎﻟﻴﺔ ﻫﻨﺩﺴﻴﺔ ﺃﺴﺎﺴﻬﺎ ‪ a‬ﻭ ﺒﺎﻟﻔﻌل ﻤﻥ ﺃﺠل ﻜل ﻋﺩﺩ ﻁﺒﻴﻌﻲ ‪ n‬ﻟﺩﻴﻨﺎ ‪:‬‬ ‫)‪Wn1 U n1  Vn1 (aU n  b)  (aVn  b‬‬ ‫‪aU n  aVn‬‬ ‫‪Wn1 a ˜Wn‬‬ ‫ﺇﺫﻥ‬‫‪ Vn1‬ﻟﺫﻟﻙ ﻨﻔﺭﺽ ﺃﻥ )‪(Vn‬‬ ‫ﻨﺭﻴﺩ ﻤﻌﺭﻓﺔ ﺍﻟﻤﺘﺘﺎﻟﻴﺔ )‪ (Vn‬ﺍﻟﺘﻲ ﺘﺤﻘﻕ ﺍﻟﻌﻼﻗﺔ ﺍﻟﺘﺭﺍﺠﻌﻴﺔ ‪aVn  b‬‬ ‫ﺜﺎﺒﺘﺔ‬ ‫ﺃﻱ ‪Vn=D‬ﻭ ﻤﻨﻪ ‪ Vn+1=D‬ﻭ ﻤﻨﻪ‬ ‫‪ Vn1 aVn  b‬ﺘﺼﺒﺢ ‪D= a D+b‬‬ ‫ﻭ ﻤﻨﻪ ‪ D - a D = b‬وﻣﻨﻪ ‪D(1-a)=b‬‬ ‫ﻋﻠﻤﺎ ﺃﻥ ‪a≠1‬‬ ‫‪α‬‬ ‫‪b‬‬ ‫ﻭ ﻨﺠﺩ ‪1  a‬‬ ‫ﻭ ﻤﻥ ﺍﻟﻌﻼﻗﺔ ‪Wn U n  Vn :‬‬ ‫‪Wn U n  α‬‬ ‫ﻨﺠﺩ‬ ‫‪Wn‬‬ ‫‪Un‬‬ ‫‬ ‫‪b‬‬ ‫‪a‬‬ ‫ﺃﻱ‬ ‫‪1‬‬ ‫‪b‬‬ ‫‪ W0‬ﺇﺫﻥ‬ ‫‪U0‬‬ ‫‬ ‫‪1 a‬‬ ‫ﺍﻷﻭل‬ ‫ﻭﺤﺩﻫﺎ‬ ‫‪a‬‬ ‫ﺃﺴﺎﺴﻬﺎ‬ ‫ﻫﻨﺩﺴﻴﺔ‬ ‫ﻤﺘﺘﺎﻟﻴﺔ‬ ‫ﻫﻲ‬ ‫)‪(Wn‬‬ ‫‪Wn‬‬ ‫‪(U 0‬‬ ‫‬ ‫‪b‬‬ ‫)‬ ‫˜‬ ‫‪a‬‬ ‫‪n‬‬ ‫‬ ‫‪1‬‬ ‫‪a‬‬ ‫ﻭ ﻟﺩﻴﻨﺎ ‪ Wn U n  α‬ﻭ ﻤﻨﻪ ‪U n Wn  α‬‬

‫‪ U n U 0 1ba ˜a n α‬‬ ‫ﻭ ﻨﺠﺩ ﺍﻟﻌﻼﻗﺔ ﺍﻷﺴﺎﺴﻴﺔ ‪:‬‬‫ﻭ ﻫﻲ ﻋﺒﺎﺭﺓ ﺍﻟﺤﺩ ﺍﻟﻌﺎﻡ ﻟﻠﻤﺘﺘﺎﻟﻴﺔ )‪ (Un‬ﺍﻟﻤﻌﺭﻓﺔ ﺒﺎﻟﻌﻼﻗﺔ ﺍﻟﺘﺭﺍﺠﻌﻴﺔ ‪U n1 aU n  b‬‬ ‫ﻤﺜﺎل ‪:‬‬ ‫ﻨﻌﺘﺒﺭ ﺍﻟﻤﺘﺘﺎﻟﻴﺔ )‪ (Un‬ﺍﻟﻤﻌﺭﻓﺔ ﻋﻠﻰ ‪ N‬ﺤﻴﺙ ‪:‬‬‫ﻁﺒﻴﻌﻲ ‪n‬‬ ‫ﺃﻋﻁ ﻋﺒﺎﺭﺓ ‪ Un‬ﺒﺩﻻﻟﺔ ‪ n‬ﻤﻥ ﺃﺠل ﻜل ﻋﺩﺩ‬ ‫‪­U‬‬ ‫‪0‬‬ ‫‪4‬‬ ‫‪¯®U‬‬ ‫‪n1‬‬ ‫‪2U n  3‬‬ ‫ﺍﻷﺠﻭﺒﺔ ‪:‬‬ ‫ﻟﺩﻴﻨﺎ ‪ b = 3 ، a = -2‬ﻭ ﻤﻨﻪ ‪:‬‬ ‫‪α‬‬ ‫‪3‬‬ ‫‪ α‬ﺃﻱ ‪1‬‬ ‫‪b‬‬ ‫)‪1  (2‬‬ ‫‪1 a‬‬ ‫ﺇﺫﻥ ﺍﻟﻤﺘﺘﺎﻟﻴﺔ )‪ (Wn‬ﺍﻟﻤﻌﺭﻓﺔ ﺒﺤﺩﻫﺎ ﺍﻟﻌﺎﻡ ‪Wn U n α‬‬‫‪ W0‬ﺇﺫﻥ ‪W0 =3‬‬ ‫ﺃﻱ ‪ Wn Un 1‬ﻫﻲ ﻤﺘﺘﺎﻟﻴﺔ ﻫﻨﺩﺴﻴﺔ ﺃﺴﺎﺴﻬﺎ ‪ a = -2‬ﻭ ﺤﺩﻫﺎ ﺍﻷﻭل ‪U 0  1‬‬ ‫ﻭ ﻤﻨﻪ ‪ Wn W0 ˜ q n‬ﺃﻱ ‪Wn = 3(-2)n‬‬ ‫ﻭ ﺒﻤﺎ ﺃﻥ ‪ U n Wn α‬ﺇﺫﻥ ‪U n 3(2) n  1‬‬ ‫‪U n1‬‬ ‫ب‪ -‬ﺇﺘﺠﺎﻩ ﺘﻐﻴﺭ ﺍﻟﻤﺘﺘﺎﻟﻴﺔ ﺍﻟﺘﺭﺍﺠﻌﻴﺔ ﺍﻟﻤﻌﺭﻓﺔ ﺒـ ‪aU n  b‬‬ ‫ﺒﻤﺎ ﺃﻥ ‪ Wn Un α‬ﺤﺴﺏ ﺍﻟﻔﺭﻉ )ﺃ(‬ ‫ﺇﺫﻥ ‪ Wn1 Un1α‬ﻭ ﻤﻨﻪ ﺇﺸﺎﺭﺓ ﺍﻟﻔﺭﻕ‬ ‫‪ U n1  U n‬ﻤﻥ ﺇﺸﺎﺭﺓ ‪Wn1  Wn‬‬ ‫ﻷﻥ ) ‪Wn1 Wn (U n1 α )(U n α‬‬ ‫‪U n1 α U n α‬‬‫‪ W0‬ﺇﺫﻥ ﺍﺘﺠﺎﻩ ﺘﻐﻴﺭ ﺍﻟﻤﺘﺘﺎﻟﻴﺔ‬ ‫‪U0 α‬‬ ‫‪U n1  U n‬‬ ‫ﻭ ﺒﻤﺎ ﺃﻥ )‪ (Wn‬ﻤﺘﺘﺎﻟﻴﺔ ﻫﻨﺩﺴﻴﺔ ﺃﺴﺎﺴﻬﺎ ‪ a‬ﻭ ﺤﺩﻫﺎ ﺍﻷﻭل‬ ‫)‪ (Un‬ﻤﻥ ﻨﻔﺱ ﺍﺘﺠﺎﻩ ﺘﻐﻴﺭ ﺍﻟﻤﺘﺘﺎﻟﻴﺔ )‪(Wn‬‬ ‫*ﻤﻼﺤﻅﺔ ‪) :‬ﺩﺭﺱ ﺍﺘﺠﺎﻫﺎﺕ ﺘﻐﻴﺭ ﺍﻟﻤﺘﺘﺎﻟﻴﺔ ﺍﻟﻬﻨﺩﺴﻴﺔ ﻓﻲ ﺩﺭﺱ ﺍﻟﻤﺘﺘﺎﻟﻴﺎﺕ ﺍﻟﻬﻨﺩﺴﻴﺔ(‬

‫ﺝ‪ -‬ﺤﺴﺎﺏ ﺍﻟﻤﺠﻤﻭﻉ ‪Sn‬‬ ‫‪Sn = U1 + U2 + …… + Un‬‬ ‫ﻟﺩﻴﻨﺎ ‪ Wn U n  α‬ﻭ ﻤﻨﻪ ‪U n Wn  α‬‬ ‫‪U1 W1  α‬‬ ‫‪U 2 W2  α‬‬ ‫‪U 3 W3  α‬‬‫) ‪(U1 U2 ... Un‬‬ ‫‪U n Wn  α‬‬ ‫‪Sn‬‬ ‫ﻭ ﺒﺎﻟﺠﻤﻊ ﻁﺭﻑ ﺇﻟﻰ ﻁﺭﻑ ﻨﺠﺩ‬ ‫‪α‬‬ ‫)‪(W1 W2 ...Wn )  (αα  ... α‬‬ ‫‪Sn‬‬ ‫ﻤﺠﻤﻭﻉ ‪ n‬ﺤﺩﺍ ﻤﻥ ﻤﺘﺘﺎﻟﻴﺔ ﻫﻨﺩﺴﻴﺔ ﺃﺴﺎﺴﻬﺎ ‪a‬‬ ‫‪ n‬ﻤﺭﺓ‬ ‫ﻭ ﻤﻨﻪ‬ ‫‪b‬‬ ‫‪ W1‬ﻭ‬ ‫ﻭ ﺤﺩﻫﺎ ﺍﻷﻭل ‪U1  α‬‬ ‫‪1 a‬‬ ‫‪ Sn W1˜(11aan )nα‬ﺇﺫﻥ ‪:‬‬ ‫‪(U‬‬ ‫‬ ‫‪b‬‬ ‫)‬ ‫‪ª1 an‬‬ ‫‪º‬‬ ‫‬ ‫‪nα‬‬ ‫‬ ‫«‬ ‫»‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪a‬‬ ‫¬‬ ‫‪1‬‬ ‫‬ ‫‪a‬‬ ‫¼‬‫ﺍﻟﺘﺭﺍﺠﻌﻴﺔ‬ ‫ﺒﺎﻟﻌﻼﻗﺔ‬ ‫ﻭ ﺘﻤﺜل ﻫﺫﻩ ﺍﻟﻤﺴﺎﻭﺍﺓ ﻋﺒﺎﺭﺓ ﺍﻟﻤﺠﻤﻭﻉ ‪ Sn‬ﻟﻠﻤﺘﺘﺎﻟﻴﺔ )‪ (Un‬ﺍﻟﻤﻌﺭﻓﺔ‬ ‫‪U n1 aU n  b‬‬ ‫ﻤﺜﺎل ‪ :‬ﻨﻌﺭﻑ ﺍﻟﻤﺘﺘﺎﻟﻴﺔ )‪ (Un‬ﻤﻥ ﺃﺠل ﻜل ﻋﺩﺩ ﻁﺒﻴﻌﻲ ‪ n‬ﻏﻴﺭ ﻤﻌﺩﻭﻤﺎ ﺒﺎﻟﺸﻜل ‪:‬‬ ‫‪­U‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪3‬‬ ‫‪®¯U‬‬ ‫‪n‬‬ ‫‪5U n  2‬‬ ‫‪ (1‬ﺃﺤﺴﺏ ‪ Un‬ﺒﺩﻻﻟﺔ ‪n‬‬ ‫‪ (2‬ﻤﺎ ﻫﻭ ﺍﺘﺠﺎﻩ ﺘﻐﻴﺭ ﺍﻟﻤﺘﺘﺎﻟﻴﺔ )‪ (Un‬؟‬ ‫‪ (3‬ﺃﺤﺴﺏ ﺍﻟﻤﺠﻤﻭﻉ ‪ Sn‬ﺍﻟﻤﻌﺭﻑ ﻜﻤﺎ ﻴﻠﻲ ‪:‬‬ ‫‪Sn = U1 + U2 + …… + Un (4‬‬

‫ﺍﻷﺠﻭﺒﺔ ‪:‬‬ ‫‪ (1‬ﺤﺴﺎﺏ ﻋﺒﺎﺭﺓ ‪ Un‬ﺒﺩﻻﻟﺔ ‪:n‬‬ ‫‪α‬‬ ‫‪b‬‬ ‫‪2‬‬ ‫‪ b = 2 ، a = 5‬ﻭﻤﻨﻪ ‪:‬‬ ‫ﻟﺩﻴﻨﺎ‬ ‫‪1 a‬‬ ‫‪15‬‬ ‫‪1‬‬ ‫‪α‬‬ ‫‬ ‫‪2‬‬ ‫ﺇﺫﻥ‬ ‫ﻨﻌﺭﻑ ﺍﻟﻤﺘﺘﺎﻟﻴﺔ )‪ (vn‬ﺒﻌﺒﺎﺭﺓ ﺤﺩﻩ ﺍﻟﻌﺎﻡ ‪Vn = Un –D :‬‬ ‫‪Vn‬‬ ‫‪Un‬‬ ‫‬ ‫‪1‬‬ ‫ﺃﻱ‬ ‫‪2‬‬ ‫ﻨﻌﻠﻡ ﺃﻥ )‪ (Vn‬ﻤﺘﺘﺎﻟﻴﺔ ﻫﻨﺩﺴﻴﺔ ﺃﺴﺎﺴﻬﺎ ‪ a=5‬ﻭ ﺤﺩﻫﺎ ﺍﻷﻭل‬ ‫‪V1‬‬ ‫‪3‬‬ ‫‬ ‫‪1‬‬ ‫‪7‬‬ ‫‪ Vn = Un – D‬أي‬ ‫‪2‬‬ ‫‪2‬‬ ‫‪Vn‬‬ ‫‪7 (5)n1‬‬ ‫و ﻣﻨﻪ ‪ Vn V1 ˜ q n1‬أي‬ ‫‪2‬‬ ‫‪ U n‬إذن ﻋﺒﺎرة ‪ Un‬ﺗﻌﻄﻰ ﺏﺎﻟﺸﻜﻞ‬ ‫‪Vn‬‬ ‫‬ ‫‪1‬‬ ‫ﻟﺪﻳﻨﺎ‬ ‫‪2‬‬ ‫‪Un‬‬ ‫‪7‬‬ ‫‪(5) n1‬‬ ‫‬ ‫‪1‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫‪ (2‬ﺍﺘﺠﺎﻩ ﺘﻐﻴﺭ )‪ (Un‬ﻤﻥ ﺍﺘﺠﺎﻩ ﺘﻐﻴﺭ )‪(Vn‬‬‫ﻭ ﺒﻤﺎ ﺃﻥ ‪ V1>0‬ﻭ ‪ q>1‬ﺇﺫﻥ )‪ (Vn‬ﻤﺘﺘﺎﻟﻴﺔ ﻤﺘﺯﺍﻴﺩﺓ ﺘﻤﺎﻤﺎ ﻋﻠﻰ ‪ N‬ﻭ ﻤﻨﻪ )‪ (Un‬ﻜﺫﻟﻙ ﻤﺘﺯﺍﻴﺩﺓ ﺘﻤﺎﻤﺎ‬ ‫ﻋﻠﻰ ‪N‬‬ ‫‪ (3‬ﺤﺴﺎﺏ ﺍﻟﻤﺠﻤﻭﻉ ‪: Sn‬‬ ‫‪Sn‬‬ ‫‪(U1‬‬ ‫‬ ‫‪α‬‬ ‫‪)(1  a n‬‬ ‫)‬ ‫‬ ‫‪nα‬‬ ‫ﻟﺩﻴﻨﺎ‬ ‫‪1 a‬‬ ‫ﻭ ﻤﻨﻪ‬‫‪Sn‬‬ ‫‪(3‬‬ ‫‬ ‫‪1 )(1  5n‬‬ ‫)‬ ‫‬ ‫‪1‬‬ ‫‪n‬‬ ‫‪2 15‬‬ ‫‪2‬‬ ‫‪Sn‬‬ ‫‪7‬‬ ‫©§¨¨‬ ‫ ‪5n‬‬ ‫‪1‬‬ ‫¸·¸‪¹‬‬ ‫‬ ‫‪1‬‬ ‫‪n‬‬ ‫‪2‬‬ ‫‪4‬‬ ‫‪2‬‬ ‫‪Sn‬‬ ‫‪7‬‬ ‫‪(5n‬‬ ‫)‪ 1‬‬ ‫‬ ‫‪1‬‬ ‫‪n‬‬ ‫ﺇﺫﻥ‬ ‫‪8‬‬ ‫‪2‬‬

‫ج‪ -‬ﺤل ﻤﺸﻜﻼﺕ ﺘﺴﺘﻌﻤل ﻓﻴﻬﺎ ﻤﺘﺘﺎﻟﻴﺎﺕ ﻤﻥ ﺍﻟﺸﻜل ‪U n1 aU n  b‬‬ ‫♦ ﻁﺭﺡ ﺍﻟﻤﺸﻜﻠﺔ ﺍﻟﺭﻴﺎﻀﻴﺔ ‪:‬‬‫ﻗ ّﺩﺭ ﺜﻤﻥ ﺴﻴﺎﺭﺓ ﺠﺩﻴﺩﺓ ﻤﻥ ﻁﺭﺍﺯ ﻤﻌﻴﻥ ﺒﻤﺒﻠﻎ ﻤﻠﻴﻭﻥ ﻭ ﺨﻤﺴﺔ ﻤﺎﺌﺔ ﺃﻟﻑ ﺩﻴﻨﺎﺭﺍ ﺠﺯﺍﺌﺭﻴﺎ ‪(1500000‬‬ ‫)‪ DA‬ﺒﺘﺎﺭﻴﺦ ‪07/01/01‬‬ ‫ﻭ ﺍﺘﻔﻕ ﻋﻠﻰ ﺒﻴﻊ ﻫﺫﺍ ﺍﻟﻨﻭﻉ ﻤﻥ ﺍﻟﺴﻴﺎﺭﺍﺕ ﺍﺒﺘﺩﺍﺀ ﻤﻥ ﻜل ﺴﻨﺔ ﺒﺎﻟﻁﺭﻴﻘﺔ ﺍﻟﺘﺎﻟﻴﺔ ‪:‬‬‫ﻜل ﻋﺎﻡ ﺍﺒﺘﺩﺍﺀ ﻤﻥ ‪ 01‬ﺠﺎﻨﻔﻲ ﺜﻤﻥ ﺍﻟﺒﻴﻊ ﺍﻟﺠﺩﻴﺩ ﻴﻨﻘﺹ ﺒﻤﻘﺩﺍﺭ ‪ 25%‬ﻤﻥ ﺜﻤﻥ ﺍﻟﺒﻴﻊ ﻟﻠﺴﻨﺔ ﺍﻟﻔﺎﺭﻁﺔ ﻭﺒﺯﻴﺎﺩﺓ‬ ‫ﻗﺩﺭﻫﺎ )‪ (50000 DA‬ﻟﻠﻤﺒﻠﻎ ﺍﻹﺠﻤﺎﻟﻲ ﺍﻟﺠﺩﻴﺩ‪.‬‬ ‫ﻨﺴﻤﻲ ‪ Pn‬ﻤﺒﻠﻎ ﺍﻟﺴﻴﺎﺭﺓ ﻴﻭﻡ ‪ 01‬ﺠﺎﻨﻔﻲ ﻤﻥ ﺴﻨﺔ ‪2007+n‬‬ ‫ﺍﻷﺴﺌﻠﺔ ‪:‬‬ ‫‪ (1‬ﺃﺤﺴﺏ ‪ P2 ، P1 ، P0‬ﺜﻡ ﺭﺘﺏ ﻫﺫﻩ ﺍﻟﺤﺩﻭﺩ‪.‬‬ ‫‪ (2‬ﺃﻜﺘﺏ ﻤﻥ ﺃﺠل ﻜل ﻋﺩﺩ ﻁﺒﻴﻌﻲ ‪ n‬ﻋﺒﺎﺭﺓ ‪ Pn+1‬ﺒﺩﻻﻟﺔ ‪Pn‬‬‫‪ (3‬ﻨﻀﻊ ﻤﻥ ﺃﺠل ﻜل ﻋﺩﺩ ﻁﺒﻴﻌﻲ ‪ Un = Pn - 200000 : n‬ﺒﺭﻫﻥ ﺃﻥ )‪ (Un‬ﻤﺘﺘﺎﻟﻴﺔ ﻫﻨﺩﺴﻴﺔ‬ ‫ﻴﻁﻠﺏ ﺘﻌﻴﻴﻥ ﺃﺴﺎﺴﻬﺎ ‪ q‬ﻭ ﺤﺩﻫﺎ ﺍﻷﻭل ‪.U0‬‬ ‫‪ (4‬ﺇﻋﻁ ﻋﺒﺎﺭﺓ ‪ Un‬ﺒﺩﻻﻟﺔ ‪ n‬ﺜﻡ ﻋﺒﺎﺭﺓ ‪ Pn‬ﺒﺩﻻﻟﺔ ‪n‬‬ ‫‪ (5‬ﺃﺤﺴﺏ ﺍﻟﻤﺠﻤﻭﻉ ‪ Sn‬ﺍﻟﻤﻌﺭﻑ ﺒـ ‪:‬‬ ‫‪Sn = P0 + P1 + … + Pn-1‬‬ ‫• ﺤل ﺍﻟﻤﺸﻜﻠﺔ ‪:‬‬ ‫‪ (1‬ﻨﻔﺭﺽ ‪ P0‬ﻫﻭ ﺜﻤﻥ ﺒﻴﻊ ﺍﻟﺴﻴﺎﺭﺓ ﻴﻭﻡ ‪07/01/01‬‬ ‫‪P0 = 1500000 DA‬‬ ‫ﺇﺫﻥ‬ ‫‪P1‬‬ ‫‪P0‬‬ ‫‬ ‫‪25‬‬ ‫‪u‬‬ ‫‪P0‬‬ ‫‬ ‫‪50000‬‬ ‫‪100‬‬ ‫‪P1‬‬ ‫ ‪P0[1‬‬ ‫]‪1‬‬ ‫‬ ‫‪50000‬‬ ‫‪4‬‬ ‫‪P1‬‬ ‫‪3‬‬ ‫‪P0‬‬ ‫‬ ‫‪50000‬‬ ‫‪4‬‬‫‪P1‬‬ ‫‪3 (1500000‬‬ ‫‪)  50000‬‬ ‫‪4‬‬ ‫‪P1 = 1175000 DA‬‬ ‫‪P2‬‬ ‫‪P1‬‬ ‫‬ ‫‪25‬‬ ‫‪P1‬‬ ‫‬ ‫‪50000‬‬ ‫‪100‬‬ ‫‪P2‬‬ ‫‪P1 (1‬‬ ‫‬ ‫‪1‬‬ ‫)‬ ‫‬ ‫‪50000‬‬ ‫‪4‬‬

‫‪P2‬‬ ‫‪3‬‬ ‫‪P1‬‬ ‫‬ ‫‪50000‬‬ ‫‪4‬‬‫‪P2‬‬ ‫‪3‬‬ ‫)‪(1175000‬‬ ‫‬ ‫‪50000‬‬ ‫‪4‬‬ ‫‪P2 = 931250 DA‬‬ ‫‪ P0‬ﻴﻤﺜل ﺜﻤﻥ ﺍﻟﺒﻴﻊ ﻴﻭﻡ ‪07/01/01‬‬ ‫‪ P1‬ﻴﻤﺜل ﺜﻤﻥ ﺍﻟﺒﻴﻊ ﻴﻭﻡ ‪08/01/01‬‬ ‫‪ P2‬ﻴﻤﺜل ﺜﻤﻥ ﺍﻟﺒﻴﻊ ﻴﻭﻡ ‪09/01/01‬‬ ‫‪P2 < P1 < P0‬‬ ‫ﻻﺤﻅ ﺃﻥ‬ ‫‪ (2‬ﻋﺒﺎﺭﺓ ‪ Pn‬ﺒﺩﻻﻟﺔ ‪n‬‬ ‫‪P1‬‬ ‫‪3‬‬ ‫‪P0‬‬ ‫‪ 50000‬‬ ‫‪:‬‬ ‫ﻟﺩﻴﻨﺎ‬ ‫‪4‬‬ ‫‪P2‬‬ ‫‪3‬‬ ‫‪P1‬‬ ‫‬ ‫‪50000‬‬ ‫‪4‬‬ ‫‪Pn1‬‬ ‫‪3‬‬ ‫‪Pn‬‬ ‫‬ ‫‪50000‬‬ ‫ﺇﺫﻥ‬ ‫‪4‬‬ ‫*ﻤﻼﺤﻅﺔ ‪ (Pn) :‬ﻤﺘﺘﺎﻟﻴﺔ ﺘﺭﺍﺠﻌﻴﺔ ﻤﻥ ﺍﻟﺸﻜل ‪Pn1 aPn  b‬‬ ‫‪b‬‬ ‫‪ a‬ﻭ ‪50000‬‬ ‫‪3‬‬ ‫ﺤﻴﺙ ‪4‬‬‫‪α‬‬ ‫‪b‬‬ ‫‪50000‬‬ ‫‪Un = Pn – 200000 (3‬‬ ‫‪1 a‬‬ ‫ﻟﺩﻴﻨﺎ ‪:‬‬ ‫‪3‬‬ ‫‪1‬‬ ‫‪4‬‬ ‫ﻭﻤﻨﻪ ‪D=200000‬‬ ‫‪D 50000 u 4‬‬‫‪ a‬ﻭ ﺤﺩﻫﺎ ﺍﻷﻭل ‪U 0 P0 α‬‬ ‫ﺇﺫﻥ ‪Un = Pn –D‬‬ ‫‪3‬‬ ‫ﺇﺫﻥ )‪ (Un‬ﻤﺘﺘﺎﻟﻴﺔ ﻫﻨﺩﺴﻴﺔ ﺃﺴﺎﺴﻬﺎ ‪ a‬ﺤﻴﺙ ‪4‬‬ ‫‪U0 = 1500000 – 200000‬‬ ‫‪U0 = 1300000‬‬

‫ﻭ ﻤﻨﻪ ‪Un = U0 . qn :‬‬ ‫¨§)‬ ‫‪3‬‬ ‫·¸‬ ‫‪n‬‬ ‫©‬ ‫‪4‬‬ ‫‪¹‬‬ ‫‪Un‬‬ ‫‪(1300000‬‬ ‫ﻭ ﺒﻤﺎ ﺃﻥ ‪Pn = Un + D‬‬‫‪Sn‬‬ ‫‪Pn‬‬ ‫) ‪(1300000‬‬ ‫‪ n‬‬‫‪3‬‬ ‫‪ (4‬ﺤﺴﺎﺏ ﺍﻟﻤﺠﻤﻭﻉ ‪ 200000‬‬ ‫‪4‬‬ ‫‪+ Pn-1‬‬ ‫… ‪Sn = P0 + P1 +‬‬ ‫ﻟﺩﻴﻨﺎ ‪ n‬ﺤﺩﺍ‪.‬‬ ‫‪Pn = Un + 200000‬‬ ‫‪P0 = U0 + 200000‬‬ ‫‪P1 = U1 + 200000‬‬ ‫‪P0  P1  }  Pn -1‬‬ ‫‪Pn-1 = Un-1 + 200000‬‬ ‫ﺒﺎﻟﺠﻤﻊ ﻁﺭﻑ ﺇﻟﻰ ﻁﺭﻑ ﻨﺠﺩ‬ ‫‪(U0 U1 }Un -1)  200000 u n‬‬ ‫ﻤﺠﻤﻭﻉ ﺤﺩﻭﺩ ﻤﺘﺘﺎﻟﻴﺔ ﻫﻨﺩﺴﻴﺔ‬ ‫‪Sn‬‬ ‫‪U‬‬ ‫‪0‬‬ ‫‪ª1  q n‬‬ ‫‪º‬‬ ‫‬ ‫‪200000‬‬ ‫‪u‬‬ ‫‪n‬‬ ‫‪:‬‬ ‫ﻤﻨﻪ‬ ‫ﻭ‬ ‫‪¬« 1  q‬‬ ‫¼»‬ ‫‪ Sn‬‬ ‫‪¬ª««114334‬‬ ‫‪n‬‬ ‫‪º‬‬ ‫»‬ ‫)‪(1300000‬‬ ‫¼»‬ ‫‪200000un‬‬ ‫‪1300000)¨§©¨1‬‬ ‫¨§‬ ‫‪3‬‬ ‫·¸‬ ‫‪n‬‬ ‫¸¸·‪¹‬‬ ‫©‬ ‫‪4‬‬ ‫‪¹‬‬ ‫‪Sn‬‬ ‫‪(4‬‬ ‫‪u‬‬ ‫‬ ‫‬ ‫‪200000‬‬ ‫‪u‬‬ ‫‪n‬‬ ‫‪(5200000)©¨¨§1‬‬ ‫¨§‬ ‫‪3‬‬ ‫·¸‬ ‫‪n‬‬ ‫·¸‬ ‫©‬ ‫‪4‬‬ ‫‪¹‬‬ ‫‪¸¹‬‬ ‫‪Sn‬‬ ‫‬ ‫‬ ‫‪200000‬‬ ‫‪u‬‬ ‫‪n‬‬ ‫*ﻤﻼﺤﻅﺔ‪:‬‬‫ﻴﻤﻜﻨﻙ ﺘﻁﺒﻴﻕ ﻗﺎﻨﻭﻥ ﺍﻟﻤﺠﻤﻭﻉ ﺍﻟﻤﺤﺼل ﻋﻠﻴﻪ ﻓﻲ ﺍﻟﺩﺭﺱ ﺍﻟﻔﻘﺭﺓ )ﺩ( ﻋﻠﻤﺎ ﺃﻥ ﺍﻟﺤﺩ ﺍﻷﻭل ﻫﻭ ‪U0‬‬

‫ﺘﻤﺎﺭﻴﻥ ﻭ ﻤﺸﻜﻼﺕ ﺤﻭل ﺍﻟﻤﺘﺘﺎﻟﻴﺎﺕ ﺍﻟﻌﺩﺩﻴﺔ‬ ‫ﺘﻤﺭﻴﻥ ‪: 1‬‬‫ﻨﻌﺘﺒﺭ ﺍﻟﻤﺘﺘﺎﻟﻴﺔ ﺍﻟﻌﺩﺩﻴﺔ )‪ (Un‬ﺍﻟﻤﻌﺭﻓﺔ ﻋﻠﻰ ‪ N‬ﺒﺤﺩﻫﺎ ﺍﻟﻌﺎﻡ ‪:Un‬‬‫‪ -‬ﺃﺤﺴﺏ ﺍﻟﺤﺩﻭﺩ ﺍﻟﺨﻤﺴﺔ ﺍﻷﻭﻟﻰ‪.‬‬‫‪ -‬ﺃﻜﺘﺏ ﻋﺒﺎﺭﺓ ﺍﻟﺤﺩ ‪ Un+1‬ﺒﺩﻻﻟﺔ ‪n.‬‬‫‪Un = 3n + 4 , Un = -2n + 1‬‬ ‫‪1‬‬ ‫‪,‬‬ ‫‪Un = 3n‬‬‫‪Un = n 1‬‬ ‫ﺘﻤﺭﻴﻥ ‪: 2‬‬‫ﻨﻌﺘﺒﺭ ﺍﻟﻤﺘﺘﺎﻟﻴﺔ )‪ (Un‬ﺍﻟﻤﻌﺭﻓﺔ ﻋﻠﻰ ‪ N‬ﺒﺎﻟﻌﻼﻗﺔ ﺍﻟﺘﺭﺍﺠﻌﻴﺔ‬‫‪ -‬ﺃﺤﺴﺏ ﺍﻟﺤﺩﻭﺩ ﺍﻟﺨﻤﺴﺔ ﺍﻷﻭﻟﻰ‪.‬‬‫‪(1) U0 = 1‬‬ ‫‪(2) U0 = -1‬‬ ‫‪Un+1 = Un + 4‬‬ ‫‪Un+1 = Un-2‬‬ ‫‪1‬‬ ‫‪(4) U0 = 2‬‬‫‪(3) U0 = 2‬‬ ‫‪Un+1 = -2Un‬‬ ‫‪Un+1 = 2Un + 1‬‬ ‫ﺘﻤﺭﻴﻥ ‪: 3‬‬‫ﻨﻌﺘﺒﺭ ﺍﻟﺩﺍﻟﺔﺍﻟﻌﺩﺩﻴﺔ ‪ f‬ﺍﻟﻤﻌﺭﻓﺔ ﻋﻠﻰ ‪ |R‬ﺒـ ‪:‬‬ ‫‪f(x) = 2x3 – 30x2 + 162‬‬ ‫ﻭﻟﻴﻜﻥ ﺤﺩﻭﺩ ﺘﻐﻴﺭﺍﺘﻬﺎ‬ ‫‪x -∞ 0 10‬‬ ‫∞‪+‬‬‫‪F(x) 162‬‬ ‫‪-838‬‬ ‫* ﺘﻌﺘﺒﺭ ﺍﻟﻤﺘﺘﺎﻟﻴﺔ ﺍﻟﻌﺩﺩﻴﺔ )‪ (Un‬ﺍﻟﻤﻌﺭﻓﺔ ﻋﻠﻰ ‪ N‬ﺒـ‪:‬‬ ‫‪Un = 2n3 – 30n2 + 162‬‬ ‫‪ (1‬ﻤﺎ ﻫﻲ ﻗﻴﻡ ‪ U10 ، U0‬؟‬‫‪ (2‬ﺇﺴﺘﻨﺘﺞ ﻤﻥ ﺠﺩﻭل ﺘﻐﻴﺭﺍﺕ ﺍﻟﺩﺍﻟﺔ ‪ f‬ﺇﺘﺠﺎﻩ ﺘﻐﻴﺭ )‪ (Un‬ﻟﻤﺎ }‪? n є { 0.1.2.3.4.5.6.7.8.9.10‬‬ ‫‪ (3‬ﺇﺴﺘﻨﺘﺞ ﻤﻥ ﺠﺩﻭل ﺘﻐﻴﺭﺍﺕ ﺍﻟﺩﺍﻟﺔ ‪ f‬ﺇﺘﺠﺎﻩ ﺘﻐﻴﺭ )‪ (Un‬ﻟﻤﺎ ‪ n ≥ 10‬؟‬

‫ﺘﻤﺭﻴﻥ ‪: 4‬‬‫ﺃﺩﺭﺱ ﺍﺘﺠﺎﻩ ﺘﻐﻴﺭ ﺍﻟﻤﺘﺘﺎﻟﻴﺔ )‪ (Un‬ﺍﻟﻤﻌﺭﻓﺔ ﻋﻠﻰ ‪ N‬ﺒﺤﺩﻫﺎ ﺍﻟﻌﺎﻡ‬‫‪Un = -2n + 5 , Un = 3n – 1 , Un = 2n‬‬ ‫‪Un‬‬ ‫=‬ ‫(‬ ‫‪1‬‬ ‫‪)n‬‬ ‫‪3‬‬ ‫ﺘﻤﺭﻴﻥ ‪: 5‬‬‫ﺤل ﺍﻟﻤﺘﺘﺎﻟﻴﺔ )‪ (Un‬ﺍﻟﻤﻌﺭﻓﺔ ﻋﻠﻰ ‪ N‬ﺤﺴﺎﺒﻴﺔ ؟‬ ‫‪(3)Un = n2‬‬‫‪(1)Un = 3n – 7 (2)Un = -4n + 2‬‬‫‪(4) U0 = -3‬‬ ‫‪(5) U0 = 1‬‬ ‫‪Un+1 = Un + n‬‬ ‫‪Un+1 = Un + 2‬‬ ‫ﺘﻤﺭﻴﻥ ‪: 6‬‬‫)‪ (Un‬ﻤﺘﺘﺎﻟﻴﺔ ﺤﺴﺎﺒﻴﺔ ﻤﻌﺭﻓﺔ ﻋﻠﻰ ‪ ،N‬ﺤﺩﻫﺎ ﺍﻷﻭل ﻫﻭ ‪ U0‬ﻭﺃﺴﺎﺴﻬﺎ ‪r‬‬ ‫ﺃﻜﺘﺏ ‪ Un‬ﺒﺩﻻﻟﺔ ‪n‬‬‫‪(1) U0 = -1 , r = 4‬‬ ‫‪3‬‬ ‫‪(2) U0 = 2 , r = -5‬‬‫‪(3) U0 =2, r = 2‬‬ ‫‪1‬‬ ‫‪(4) U0 = 4 , r = 2‬‬ ‫ﺘﻤﺭﻴﻥ ‪: 7‬‬‫)‪ (Un‬ﻤﺘﺘﺎﻟﻴﺔ ﺤﺴﺎﺒﻴﺔ ﻤﻌﺭﻓﺔ ﻋﻠﻰ ‪ N‬ﺃﺴﺎﺴﻬﺎ ‪ r‬ﺒﺤﻴﺙ ‪:‬‬ ‫‪U5 = 9 , U2 = 3‬‬ ‫‪ (1‬ﺃﺤﺴﺏ ﺍﻷﺴﺎﺱ ‪.r‬‬ ‫‪ (2‬ﺃﺤﺴﺏ ﺍﻟﺤﺩ ﺍﻷﻭل ‪.U0‬‬ ‫‪ (3‬ﺃﻜﺘﺏ ‪ Un‬ﺒﺩﻻﻟﺔ ‪n‬‬‫‪ (4‬ﺃﺤﺴﺏ ﺍﻟﻤﺠﻤﻭﻉ ‪Sn = U0 + U1 + …+Un :Sn‬‬ ‫‪ (5‬ﺍﺴﺘﻨﺘﺞ ﺍﻟﻤﺠﻤﻭﻉ ‪.S10 , S20‬‬ ‫‪ -‬ﻨﻔﺱ ﺍﻷﺴﺌﻠﺔ ﺇﺫﺍ ﻜﺎﻥ‪U8 = 2 , U4 = 10 :‬‬

‫ﺘﻤﺭﻴﻥ ‪: 8‬‬ ‫)‪ (Un‬ﻤﺘﺘﺎﻟﻴﺔ ﺤﺴﺎﺒﻴﺔﻤﻌﺭﻓﺔ ﻋﻠﻰ ‪، N‬ﺍﺴﺎﺴﻬﺎ ‪ ( r ¢ 0 ) r‬ﺒﺤﻴﺙ ‪:‬‬ ‫‪ U3+U4+U5=42‬ﻭ ‪(U3)2+(U4)2+(U5 )2=381‬‬ ‫‪ (1‬ﺃﺤﺴﺏ ﺍﻟﺤﺩﻭﺩ ‪U3,U4,U5‬‬ ‫‪ (2‬ﺃﺤﺴﺏ ﺍﻷﺴﺎﺱ ﺜﻡ ﺍﻟﺤﺩ ﺍﻷﻭل‬ ‫‪ ( 3‬ﺃﻜﺘﺏ ‪ Un‬ﺒﺩﻻﻟﺔ ‪n‬‬ ‫ﺘﻤﺭﻴﻥ‪: 9‬‬ ‫)‪ (Un‬ﻤﺘﺘﺎﻟﻴﺔ ﺤﺴﺎﺒﻴﺔﻤﻌﺭﻓﺔ ﻋﻠﻰ ‪، N‬ﺍﺴﺎﺴﻬﺎ ‪ r‬ﺒﺤﻴﺙ ‪:‬‬ ‫‪ r = 3‬ﻭ ‪U1 + U2 + U3 + U4 = 34‬‬ ‫‪ (1‬ﺃﺤﺴﺏ ‪U0‬‬ ‫‪ (2‬ﺃﻜﺘﺏ ‪ Un‬ﺒﺩﻻﻟﺔ ‪n‬‬ ‫‪ (3‬ﺃﺤﺴﺏ‪Sn = U0+…+Un :‬‬ ‫‪ (4‬ﺃﺤﺴﺏ ﺍﻟﻤﺠﻤﻭﻉ‪S'n = U1+…+Un :‬‬ ‫ﺘﻤﺭﻴﻥ ‪: 10‬‬ ‫ﻜﺎﻥ ﻋﺩﺩ ﻋﻤﺎل ﻤﺅﺴﺴﺔ ﺇﻨﺘﺎﺠﻴﺔ ﻋﺎﻡ ‪ 2000‬ﻫﻭ ‪ 300‬ﻋﺎﻤل‬ ‫ﺒﺤﻴﺙ ﺘﻨﻁﻕ ﺍﻟﻤﺅﺴﺴﺔ ﻜل ﺴﻨﺔ ‪ 40‬ﻋﺎﻤﻼ ﺠﺩﻴﺩﺍ‪.‬‬ ‫ﻨﻌﺘﺒﺭ ﺃﻥ‪ V0‬ﻫﻭ ﻋﺩﺩ ﺍﻟﻌﻤﺎل ﻋﺎﻡ ‪2000‬‬ ‫ﻭ‪ vn‬ﻫﻭ ﻋﺩﺩ ﺍﻟﻌﻤﺎل ﺒﻌﺩ ‪ n‬ﺴﻨﺔ‬ ‫‪ -1‬ﺃﺤﺴﺏ ﻋﺩﺩ ﺍﻟﻌﻤﺎل ﻋﺎﻡ ‪.2003 ، 2002 ، 2001‬‬ ‫‪ -2‬ﺃﻭﺠﺩ ﻋﻼﻗﺔ ﺒﻴﻥ ‪ vn+1‬ﻭ ‪vn‬‬ ‫‪ -3‬ﺍﺴﺘﻨﺘﺞ ﻁﺒﻴﻌﺔ ﺍﻟﻤﺘﺘﺎﻟﻴﺔ )‪(vn‬‬ ‫‪ -4‬ﺃﻜﺘﺏ ‪ vn‬ﺒﺩﻻﻟﺔ ‪n.‬‬ ‫‪ – 5‬ﺃﺤﺴﺏ ﺍﻟﻤﺠﻤﻭﻉ ‪Sn=V0 +V1 +…+Vn :‬‬ ‫ﺘﻤﺭﻴﻥ ‪: 11‬‬‫ﻭﻀﻊ ﺸﺨﺹ ﻤﺒﻠﻎ ‪ 2000‬ﺩ‪.‬ﺝ ﻋﺎﻡ ‪ 2007‬ﺒﺄﺤﺩ ﺍﻟﺒﻨﻭﻙ‪ ،‬ﺒﺤﻴﺙ ﻟﻪ ﻓﺎﺌﺩﺓ ﺒﺴﻴﻁﺔ ﺴﻨﻭﻴﺔ ﻗﺩﺭﻫﺎ ‪. %5‬‬ ‫ﻨﻔﺭﺽ ﺃﻥ ﺍﻟﻤﺒﻠﻎ ﺍﻟﺫﻱ ﻭﻀﻌﻪ ﻋﺎﻡ ‪ 2007‬ﻫﻭ ‪U0‬‬

‫ﻭ ‪ : Un‬ﺍﻟﺭﺼﻴﺩ ﺍﻟﻤﺤﺼل ﻋﻠﻴﻪ ﺒﻌﺩ ‪ n‬ﺴﻨﻭﺍﺕ‬ ‫‪ (1‬ﺃﺤﺴﺏ ﺭﺼﻴﺩ ﺍﻟﺸﺨﺹ ﻋﺎﻡ ‪.2010 ،2009 ، 2008‬‬ ‫‪ (2‬ﺃﻭﺠﺩ ﻋﻼﻗﺔ ﺒﻴﻥ ‪ Un+ 1‬ﻭ ‪Un‬‬ ‫‪ (3‬ﺍﺴﺘﻨﺘﺞ ﻁﺒﻴﻌﺔ )‪(Un‬‬ ‫‪ (4‬ﺃﻜﺘﺏ ‪ Un‬ﺒﺩﻻﻟﺔ ‪n‬‬ ‫ﺘﻤﺭﻴﻥ ‪: 12‬‬ ‫)‪ (Un‬ﻤﺘﺘﺎﻟﻴﺔ ﺤﺴﺎﺒﻴﺔ ﻤﻌﺭﻓﺔ ﻋﻠﻰ ‪N‬‬‫ﺃﺴﺎﺴﻬﺎ ‪ r‬ﻭﺤﺩﻫﺎ ﺍﻷﻭل ‪ ، U0 = 1‬ﻴﻌﺘﺒﺭ ﺍﻟﻤﺠﻤﻭﻉ ‪ S‬ﻟﻌﺩﺓ ﺤﺩﻭﺩ ‪:‬‬ ‫‪S = 1+11+21+…+201‬‬ ‫‪ (1‬ﺃﺤﺴﺏ ﺍﻷﺴﺎﺱ ‪r‬‬ ‫‪ (2‬ﺃﻜﺘﺏ ‪ Un‬ﺒﺩﻻﻟﺔ ‪n‬‬ ‫‪ (3‬ﻋﻴﻥ ﺍﻟﺭﺘﺒﺔ ‪ n‬ﺒﺤﻴﺙ ‪U1 = 201 :‬‬ ‫‪ (4‬ﻨﻌﺘﺒﺭ ﺍﻟﻤﺠﻤﻭﻉ ‪Sn = U0+…+Un : Sn‬‬ ‫ﻫل ﺘﻭﺠﺩ ﻗﻴﻤﺔ ﻟـ‪ n‬ﺒﺤﻴﺙ ‪Sn = 105 :‬‬ ‫ﺘﻤﺭﻴﻥ ‪: 13‬‬ ‫ﻫل ﺍﻟﻤﺘﺘﺎﻟﻴﺔ )‪ (Un‬ﺍﻟﻤﻌﺭﻓﺔ ﻋﻠﻰ ‪ N‬ﻫﻨﺩﺴﻴﺔ ؟‬ ‫‪Un = 3.(2)n , Un = (-4).(3)n , Un = n2‬‬ ‫ﺘﻤﺭﻴﻥ ‪: 14‬‬ ‫)‪ (Un‬ﻤﺘﺘﺎﻟﻴﺔ ﻫﻨﺩﺴﻴﺔ ﺃﺴﺎﺴﻬﺎ ‪ q‬ﻭﺤﺩﻫﺎ ﺍﻷﻭل ‪U0‬‬ ‫‪(1‬ﺃﻜﺘﺏ ‪ Un‬ﺒﺩﻻﻟﺔ ‪n‬‬ ‫‪ (2‬ﺍﺩﺭﺱ ﺍﺘﺠﺎﻩ ﺘﻐﻴﺭ )‪(Un‬‬ ‫‪q = 2 , U0 = 3‬‬ ‫‪1‬‬ ‫‪q = ( 3 ) , U0 = 2‬‬ ‫‪1‬‬ ‫‪q = (-2) , U0 = 2‬‬

‫ﺘﻤﺭﻴﻥ ‪: 15‬‬‫)‪ (Un‬ﻤﺘﺘﺎﻟﻴﺔ ﻫﻨﺩﺴﻴﺔﻤﻌﺭﻓﺔ ﻋﻠﻰ‪ *N‬ﺃﺴﺎﺴﻬﺎ ‪ q‬ﻭﺤﺩﻫﺎ ﺍﻷﻭل‪U1‬‬‫ﻨﻔﺱ ﺃﺴﺌﻠﺔ ﺍﻟﺘﻤﺭﻴﻥ ‪14‬‬ ‫‪q = 2 , U1 = -3‬‬‫‪q‬‬ ‫=‬ ‫(‬ ‫‪1‬‬ ‫)‬ ‫‪,‬‬ ‫‪U1‬‬ ‫=‬ ‫‪4‬‬ ‫‪2‬‬ ‫‪q = -3 , U1 = + 2‬‬ ‫ﺘﻤﺭﻴﻥ ‪: 16‬‬ ‫)‪ (Un‬ﻤﺘﺘﺎﻟﻴﺔ ﻫﻨﺩﺴﻴﺔ ﻤﻌﺭﻓﺔ ﻋﻠﻰ ‪ N‬ﺒﺤﻴﺙ ‪:‬‬ ‫‪U4 = 12 , U2 = 3‬‬ ‫ﻭﺃﺴﺎﺴﻬﺎ ‪(q > 0) q‬‬ ‫‪ (1‬ﺃﺤﺴﺏ ﺍﻷﺴﺎﺱ ‪q‬‬ ‫‪ (2‬ﺃﺤﺴﺏ ﺍﻟﺤﺩ ﺍﻷﻭل ‪U0‬‬ ‫‪ (3‬ﺃﻜﺘﺏ ‪ Un‬ﺒﺩﻻﻟﺔ ‪n‬‬ ‫‪ (4‬ﺃﺤﺴﺏ ﺍﻟﻤﺠﻤﻭﻉ ‪Sn = U0+…+Un : Sn‬‬ ‫ﺍﺴﺘﻨﺘﺞ ﺍﻟﻤﺠﻤﻭﻉ ‪S6‬‬ ‫ﺘﻤﺭﻴﻥ ‪: 17‬‬‫)‪ (Un‬ﻤﺘﺘﺎﻟﻴﺔ ﻫﻨﺩﺴﻴﺔ ﻤﻌﺭﻓﺔ ﻋﻠﻰ *‪U5 = 54 , U3 = 6 ، N‬‬ ‫ﻭﺃﺴﺎﺴﻬﺎ ‪(q > 0) q‬‬ ‫‪ (1‬ﺃﺤﺴﺏ ﺍﻷﺴﺎﺱ ‪q‬‬ ‫‪ (2‬ﺃﺤﺴﺏ ﺍﻟﺤﺩ ﺍﻷﻭل ‪U1‬‬ ‫‪ (3‬ﺃﻜﺘﺏ ‪ Un‬ﺒﺩﻻﻟﺔ ‪n‬‬ ‫‪ (4‬ﺃﺤﺴﺏ ﺍﻟﻤﺠﻤﻭﻉ ‪Sn = U1+…+Un : Sn‬‬ ‫ﺜﻡ ﺍﺴﺘﻨﺘﺞ ‪S6‬‬ ‫ﺘﻤﺭﻴﻥ ‪: 18‬‬ ‫)‪ (Un‬ﻤﺘﺘﺎﻟﻴﺔ ﻫﻨﺩﺴﻴﺔ ﻤﻌﺭﻓﺔ ﻋﻠﻰ ‪N‬‬ ‫ﺃﺴﺎﺴﻬﺎ )‪ ( q > 0‬ﻭ ﺤﺩﻫﺎ ﺍﻷﻭل ‪U0 = 1‬‬ ‫ﺒﺤﻴﺙ ‪U0 + U1 + U2 = 13 :‬‬ ‫‪ (1‬ﺃﺤﺴﺏ ﺍﻷﺴﺎﺱ ‪q‬‬ ‫‪ (2‬ﺃﻜﺘﺏ ‪ Un‬ﺒﺩﻻﻟﺔ ‪n‬‬

‫ﺘﻤﺭﻴﻥ ‪: 19‬‬‫ﺃﻭﺩﻉ ﺸﺨﺹ ﻤﺒﻠﻐﺎ ﻗﺩﺭﻩ ‪11000‬ﺩ‪.‬ﺝ ﺒﺈﺤﺩﻯ ﺍﻟﺒﻨﻭﻙ ﻋﺎﻡ ‪ 2000‬ﺒﺤﻴﺙ ﺤﺼل ﻟﻪ ﻓﺎﺌﺩﺓ ﺴﻨﻭﻴﺔ ﻤﺭﻜﺒﺔ‬ ‫ﻗﺩﺭﻫﺎ ‪.% 6‬‬ ‫ﺇﺫﺍ ﺇﻋﺘﺒﺭﻨﺎ ﺃﻥ ﺍﻟﻤﺒﻠﻎ ﺍﻟﻤﻭﺩﻉ ﻫﻭ ‪vn‬‬ ‫ﻭﻨﻌﺘﺒﺭ ﺍﻟﻌﺩﺩ ‪ : vn‬ﺍﻟﺭﺼﻴﺩ ﺍﻟﺠﺩﻴﺩ ﺒﻌﺩ ‪ n‬ﺴﻨﻭﺍﺕ‬ ‫‪ (1‬ﺃﺤﺴﺏ ﺍﻟﻤﺒﻠﻎ ﺍﻟﻤﺤﺼل ﻋﻠﻴﻪ ﻋﺎﻡ ‪2003 ، 2002 ، 2001‬‬ ‫‪ (2‬ﺃﻭﺠﺩ ﻋﻼﻗﺔ ﺒﻴﻥ ‪ vn+ 1‬ﻭ ‪vn‬‬ ‫‪ (3‬ﺍﺴﺘﻨﺘﺞ ﻁﺒﻴﻌﺔ ﺍﻟﻤﺘﺘﺎﻟﻴﺔ) ‪(Vn‬‬ ‫‪ (4‬ﺃﻜﺘﺏ ‪ Vn‬ﻟﺩﻻﻟﺔ ‪n‬‬ ‫ﺘﻤﺭﻴﻥ ‪: 20‬‬ ‫ﺇﻨﺘﺎﺝ ﻤﺼﻨﻊ ﻋﺎﻡ‪ 2006‬ﻫﻭ ‪ 3000‬ﻁﻥ‪ ،‬ﻭ ﻴﺯﻴﺩ ﺴﻨﻭﻴﺎ ﺒـ ‪2%‬‬ ‫ﺇﺫﺍ ﺍﻋﺘﺒﺭﻨﺎ ‪ V0‬ﻫﻭ ﺍﻹﻨﺘﺎﺝ ﻋﺎﻡ ‪2006‬‬ ‫ﻭ ‪ :Vn‬ﺍﻹﻨﺘﺎﺝ ﺒﻌﺩ ‪ n‬ﺴﻨﺔ‬ ‫‪ (1‬ﺃﺤﺴﺏ ﺍﻹﻨﺘﺎﺝ ﻋﺎﻡ ‪. 2008 ، 2007‬‬ ‫‪ (2‬ﺃﻭﺠﺩ ﻋﻼﻗﺔ ﺒﻴﻥ ‪ vn+ 1‬ﻭ ‪vn‬‬ ‫‪ (3‬ﺍﺴﺘﻨﺘﺞ ﻁﺒﻴﻌﺔ ﺍﻟﻤﺘﺘﺎﻟﻴﺔ )‪(Vn‬‬ ‫‪ (4‬ﺃﻜﺘﺏ ‪ Vn‬ﻟﺩﻻﻟﺔ ‪n‬‬ ‫‪Sn = V0+…+Vn‬‬ ‫‪ (5‬ﺃﺤﺴﺏ ﺍﻟﻤﺠﻤﻭﻉ ‪:‬‬ ‫ﺘﻤﺭﻴﻥ ‪: 21‬‬ ‫‪ (1‬ﻨﻌﺘﺒﺭ ﺍﻟﻤﺘﺘﺎﻟﻴﺔ )‪ (Un‬ﺍﻟﻤﻌﺭﻓﺔ ﻋﻠﻰ‪N‬ﺒﺎﻟﻌﻼﻗﺔ ﺍﻟﺘﺭﺍﺠﻌﻴﺔ‬ ‫‪U0 = 1‬‬ ‫‪2 Un+1 = -3 Un +‬‬ ‫ﺃﺤﺴﺏ ‪U1 ,U2 ,U3‬‬‫‪Vn‬‬ ‫‪U‬‬ ‫‪n‬‬ ‫‬ ‫‪1‬‬ ‫ﻨﻌﺘﺒﺭ ﺍﻟﻤﺘﺘﺎﻟﻴﺔ )‪ (Vn‬ﺍﻟﻤﻌﺭﻓﺔ ﻋﻠﻰ ‪ N‬ﺒـ‪:‬‬ ‫‪(2‬‬ ‫‪2‬‬ ‫ﺍ‪ /‬ﺃﺤﺴﺏ ‪V2, V1,V0 :‬‬ ‫ﺝ‪ /‬ﺒﻴﻥ ﺃﻥ)‪ (Vn‬ﻫﻲ ﻤﺘﺘﺎﻟﻴﺔ ﻫﻨﺩﺴﻴﺔ ﺃﺴﺎﺴﻬﺎ‪-3‬‬ ‫‪ (3‬ﺃﻜﺘﺏ ‪ Vn‬ﺒﺩﻻﻟﺔ ‪n‬‬ ‫‪ (4‬ﺃﻜﺘﺏ‪ Un‬ﺩﻻﻟﺔ ‪n‬‬ ‫‪Sn=V0+..+Vn‬‬ ‫‪ (5‬ﺃﺤﺴﺏ ﺍﻟﻤﺠﻤﻭﻉ‬ ‫‪Snc U0 ...U n‬‬ ‫ﺜﻡ ﺃﺤﺴﺏ ﺍﻟﻤﺠﻤﻭﻉ ‪:‬‬

‫ﺤـﻠﻭل ﺍﻟﺘﻤﺎﺭﻴﻥ ‪:‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ ‪: 1‬‬ ‫‪Un = 3n + 4‬‬ ‫‪(1‬‬ ‫ﺤﺴﺎﺏ ﺍﻟﺤﺩﻭﺩ ﺍﻟﺨﻤﺴﺔ ﺍﻷﻭﻟﻰ‪:‬‬ ‫ﻤﻥ ﺃﺠل ‪ n = 0‬ﻓﺈﻥ‪ U0 = 3(0)+4 = 4 :‬ﺇﺫﻥ ‪U0 = 4‬‬ ‫ﻤﻥ ﺃﺠل ‪ n = 1‬ﻓﺈﻥ‪ U1 = 3(1)+4 = 7 :‬ﺇﺫﻥ ‪U1 = 7‬‬ ‫ﻤﻥ ﺃﺠل ‪ n = 2‬ﻓﺈﻥ‪ U2 = 3(2)+4 = 10 :‬ﺇﺫﻥ ‪U2 = 10‬‬ ‫ﻤﻥ ﺃﺠل ‪ n = 3‬ﻓﺈﻥ‪ U3 = 3(3)+4 = 13 :‬ﺇﺫﻥ ‪U3 = 13‬‬ ‫ﻤﻥ ﺃﺠل ‪ n = 4‬ﻓﺈﻥ‪ U4 = 3(4)+4 = 16 :‬ﺇﺫﻥ ‪U4 = 16‬‬ ‫‪ (2‬ﻜﺘﺎﺒﺔ ‪ Un+1‬ﺒﺩﻻﻟﺔ ‪:n‬‬‫ﻓﻲ ﻋﺒﺎﺭﺓ ﺍﻟﺤﺩ ﺍﻟﻌﺎﻡ ‪ Un‬ﻟﻠﺤﺼﻭل ﻋﻠﻰ ﻋﺒﺎﺭﺓ ‪ Un+1‬ﻨﻘﻭﻡ ﺒﺘﻌﻭﻴﺽ ﺍﻟ ّﺭﺘﺒﺔ ‪ n‬ﺒﺎﻟ ّﺭﺘﺒﺔ )‪ (n+1‬ﻓﻨﺠﺩ‪:‬‬ ‫‪Un+1 = 3(n+1) + 4‬‬ ‫ﻨﻘﻭﻡ ﺒﺎﻟﻨﺸﺭ ﻓﻨﺠﺩ‪:‬‬ ‫‪Un+1 = 3n + 3 + 4‬‬ ‫ﺇﺫﻥ‪Un+1 = 3n + 7 :‬‬ ‫‪1‬‬ ‫‪U0‬‬ ‫=‬ ‫‪1‬‬ ‫=‬ ‫‪1‬‬ ‫=‬ ‫‪1‬‬ ‫ﻤﻥ ﺃﺠل ‪ n = 0‬ﻓﺈﻥ‪:‬‬‫‪2) Un = n  1‬‬ ‫‪0 1‬‬ ‫‪1‬‬ ‫ﻤﻥ ﺃﺠل ‪ n = 1‬ﻓﺈﻥ‪:‬‬ ‫ﻤﻥ ﺃﺠل ‪ n = 2‬ﻓﺈﻥ‪:‬‬ ‫ﺇﺫﻥ‪U0 = 1 :‬‬ ‫ﻤﻥ ﺃﺠل ‪ n = 3‬ﻓﺈﻥ‪:‬‬ ‫ﻤﻥ ﺃﺠل ‪ n = 4‬ﻓﺈﻥ‪:‬‬‫ﺇﺫﻥ‪U1 = 1 :‬‬ ‫= ‪U1 = 1‬‬ ‫‪1‬‬ ‫‪2‬‬ ‫‪11‬‬ ‫‪2‬‬‫‪U2‬‬ ‫=‬ ‫‪1‬‬ ‫ﺇﺫﻥ‪:‬‬ ‫‪U2‬‬ ‫=‬ ‫‪2‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫=‬ ‫‪1‬‬ ‫‪3‬‬ ‫‬ ‫‪3‬‬‫ﺇﺫﻥ‪U3 = 1 :‬‬ ‫= ‪U3 = 1‬‬ ‫‪1‬‬ ‫‪4‬‬ ‫‪4‬‬ ‫‪31‬‬‫‪U4‬‬ ‫=‬ ‫‪1‬‬ ‫ﺇﺫﻥ‪:‬‬ ‫‪U4‬‬ ‫=‬ ‫‪4‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫=‬ ‫‪1‬‬ ‫‪5‬‬ ‫‬ ‫‪5‬‬

‫‪ (2‬ﻋﺒﺎﺭﺓ ﺍﻟﺤﺩ ‪ Un+1‬ﺒﺩﻻﻟﺔ ‪: n‬‬ ‫ﻟﺩﻴﻨﺎ‪ :‬ﺒﺘﻌﻭﻴﺽ ﺍﻟﺭﺘﺒﺔ ‪ n‬ﺒﺎﻟﺭﺘﺒﺔ )‪(n+1‬‬‫‪Un+1‬‬ ‫=‬ ‫‪1‬‬ ‫=‬ ‫‪n‬‬ ‫‪1‬‬ ‫‬ ‫‪1‬‬ ‫=‬ ‫‪1‬‬ ‫ﻨﺠﺩ ‪:‬‬ ‫)‪ 1‬‬ ‫‪1‬‬ ‫‪n2‬‬ ‫‪(n‬‬ ‫‬ ‫‪1‬‬ ‫‪1‬‬ ‫ﺇﺫﻥ‪Un+1 = n  2 :‬‬ ‫‪Un = 3n (3‬‬ ‫‪ -‬ﺤﺴﺎﺏ ﺍﻟﺤﺩﻭﺩ ﺍﻟﺨﻤﺴﺔ ﺍﻷﻭﻟﻰ‪:‬‬ ‫ﻤﻥ ﺃﺠل ‪ n = 0‬ﻓﺈﻥ‪U0 = 30 = 1 :‬‬ ‫ﻤﻥ ﺃﺠل ‪ n = 1‬ﻓﺈﻥ‪U1 = 31 = 3 :‬‬ ‫ﻤﻥ ﺃﺠل ‪ n = 2‬ﻓﺈﻥ‪U2 = 32 = 9 :‬‬ ‫ﻤﻥ ﺃﺠل ‪ n = 3‬ﻓﺈﻥ‪U3 = 33 = 27 :‬‬ ‫ﻤﻥ ﺃﺠل ‪ n = 4‬ﻓﺈﻥ‪U4 = 34 = 81 :‬‬ ‫‪ -‬ﻜﺘﺎﺒﺔ ﺍﻟﺤﺩ ‪ Un+1‬ﺒﺩﻻﻟﺔ ‪:n‬‬ ‫ﻟﺩﻴﻨﺎ‪Un+1 = 3(n+1) = 3n x 3 = 3 x 3n :‬‬ ‫ﺇﺫﻥ‪Un+1 = 3 x 3n :‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ ‪: 2‬‬ ‫‪1‬‬‫‪(1) U0 = 2‬‬ ‫‪Un+1 = 2Un + 1‬‬ ‫ﺤﺴﺎﺏ ﺍﻟﺤﺩﻭﺩ‪U1 , U2 , U3 , U4 , U5 :‬‬‫ﻓﻲ ﺍﻟﻤﺘﺘﺎﻟﻴﺔ ﺍﻟﻤﻌﺭﻓﺔ ﺒﺎﻟﻌﻼﻗﺔ ﺍﻟﺘﺭﺍﺠﻌﻴﺔ‪ ،‬ﺤﺴﺎﺏ ﻜل ﺤﺩ ﻤﺘﻌﻠﻕ ﺒﺎﻟﺤﺩ ﺍﻟ ّﺴﺎﺒﻕ ﻟﻪ‪ ،‬ﺃﻱ ﻟﺤﺴﺎﺏ ‪،U1‬‬ ‫ﻴﺠﺏ ﺃﻥ ﻨﻌﻠﻡ ﻗﻴﻤﺔ ﺍﻟﺤﺩ ‪.U0‬‬ ‫ﻭﻟﺤﺴﺎﺏ ﺍﻟﺤﺩ ‪ ،U2‬ﻴﺠﺏ ﺃﻥ ﻨﻌﻠﻡ ﻗﻴﻤﺔ ﺍﻟﺤﺩ ‪.U1‬‬ ‫* ﺤﺴﺎﺏ ‪ : U1‬ﻟﺩﻴﻨﺎ ﺒﺘﻌﻭﻴﺽ ﻗﻴﻤﺔ ‪ n‬ﺒـ ‪ 0‬ﻓﻲ ﺍﻟﻌﻼﻗﺔ ﺍﻟﺘﺭﺍﺠﻌﻴﺔ ﻨﺠﺩ ‪:‬‬ ‫‪U0+1 = 2U0 + 1‬‬

‫‪1‬‬ ‫ﻭﻤﻨﻪ‪ U1 = 2U0 + 1 :‬ﺇﺫﻥ‪U1 = 2( 2 ) + 1 :‬‬ ‫ﻭﻤﻨﻪ‪U1 = 2 U1 = 1+ 1 :‬‬ ‫* ﺤﺴﺎﺏ ‪ : U2‬ﻨﻌﻭﺽ ﻗﻴﻤﺔ ‪ n‬ﺒﺎﻟﻌﺩﺩ ‪ 1‬ﻨﺠﺩ ‪:‬‬ ‫‪ U1+1 = 2U1 + 1‬ﻭﻤﻨﻪ‪ U2 = 2U1 + 1 :‬ﻭﻤﻨﻪ‪U2 = 2(2) + 1 :‬‬ ‫ﺇﺫﻥ‪U2 = 5 :‬‬ ‫* ﺤﺴﺎﺏ ‪ : U3‬ﻨﻌﻭﺽ ﻗﻴﻤﺔ ‪ n‬ﺒﺎﻟﻌﺩﺩ ‪ 2‬ﻨﺠﺩ ‪:‬‬ ‫‪ U2+1 = 2U2 + 1‬ﻭﻤﻨﻪ‪ U3 = 2U2 + 1 :‬ﻭﻤﻨﻪ ‪U3 = 2(5) + 1 :‬‬ ‫ﺇﺫﻥ‪U3 = 11 :‬‬ ‫* ﺤﺴﺎﺏ ‪ : U4‬ﻨﻌﻭﺽ ﻗﻴﻤﺔ ‪ n‬ﺒﺎﻟﻌﺩﺩ ‪ 3‬ﻨﺠﺩ‪:‬‬‫‪ U3+1 = 2U3 + 1‬ﻭﻤﻨﻪ‪ U4 = 2U3 + 1 :‬ﻭﻤﻨﻪ‪U4 = 2(11) + 1 :‬‬ ‫ﺇﺫﻥ‪23U4 = :‬‬ ‫* ﺤﺴﺎﺏ ‪ : U5‬ﻨﻌﻭﺽ ﻗﻴﻤﺔ ‪ n‬ﺒﺎﻟﻌﺩﺩ ‪ 4‬ﻨﺠﺩ‪:‬‬‫‪ U4+1 = 2U4 + 1‬ﻭﻤﻨﻪ‪ U5 = 2U4 + 1 :‬ﻭﻤﻨﻪ‪U5 = 2(23) + 1 :‬‬ ‫ﺇﺫﻥ‪U5 = 45 :‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ ‪: 3‬‬‫‪ f(x) = 2x3 – 30 x2 + 162‬ﻭﺍﻟﺘﺎﻟﻲ ﺠﺩﻭل ﺘﻐﻴﺭﺍﺕ ‪: f‬‬ ‫‪x -∞ 0 10‬‬ ‫∞‪+‬‬‫‪F(x) 162‬‬ ‫‪-838‬‬ ‫ﻭﻟﺘﻜﻥ ﺍﻟﻤﺘﺘﺎﻟﻴﺔ )‪ (Un‬ﺍﻟﻤﻌﺭﻓﺔ ﻋﻠﻰ ‪:N‬‬ ‫‪Un = 2n3 – 30n2 + 162‬‬ ‫ﺇﻴﺠﺎﺩ ﻗﻴﻡ ‪ U10 , U0‬؟‬ ‫ﻤﻥ ﺠﺩﻭل ﺍﻟﺘﻐﻴﺭﺍﺕ ﻟﺩﻴﻨﺎ‪U0 = 162 :‬‬ ‫‪U10 = -838‬‬ ‫‪ (2‬ﺍﺘﺠﺎﻩ ﺘﻐﻴﺭ )‪ (Un‬ﻟﻤﺎ }‪n є { 0.1.2.3.4.5.6.7.8.9.10‬؟‬‫ﺍﺘﺠﺎﻩ ﺘﻐﻴﺭ )‪ (Un‬ﻋﻠﻰ ﻫﺫﻩ ﺍﻟﻤﺠﻤﻭﻋﺔ‪ ،‬ﻫﻭ ﻨﻔﺱ ﺍﺘﺠﺎﻩ ﺘﻐﻴﺭ ﺍﻟﺩﺍﻟﺔ ‪ f‬ﻋﻠﻰ ﺍﻟﻤﺠﺎل ]‪[0 , 10‬‬ ‫ﻭﻤﻥ ﺠﺩﻭل ﺘﻐﻴﺭﺍﺕ ﺍﻟﺩﺍﻟﺔ ‪ ،f‬ﻭﺍﻀﺢ ﺃﻥ ‪ f‬ﻤﺘﻨﺎﻗﺼﺔ ﻋﻠﻰ ﺍﻟﻤﺠﺎل‬ ‫]‪ [0 , 10‬ﺇﺫﻥ )‪ (Un‬ﻤﺘﻨﺎﻗﺼﺔ ﻋﻠﻰ ﺍﻟﻤﺠﻤﻭﻋﺔ }‪{0.1.2.3.4.5.6.7.8.9.10‬‬ ‫‪ (3‬ﺍﺘﺠﺎﻩ ﺘﻐﻴﺭ )‪ (Un‬ﻟﻤﺎ ‪ n ≥ 10‬ﻫﻭ ﻨﻔﺱ ﺍﺘﺠﺎﻩ ﺘﻐﻴﺭ ﺍﻟﺩﺍﻟﺔ ‪ f‬ﻋﻠﻰ‬

‫[∞‪ [10 , +‬ﺇﺫﻥ )‪ (Un‬ﻤﺘﺯﺍﻴﺩﺓ ﻟﻤﺎ ‪n ≥ 10‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ ‪: 4‬‬ ‫‪Un = -2n + 5 (1‬‬ ‫ﺩﺭﺍﺴﺔ ﺍﺘﺠﺎﻩ )‪(Un‬‬ ‫ﺃﻭﻻ‪ :‬ﻨﻜﺘﺏ ﺍﻟﺤﺩ ‪ Un+1‬ﺒﺩﻻﻟﺔ ‪.n‬‬‫ﻟﻜﺘﺎﺒﺔ ﺍﻟﺤﺩ ‪ Un+1‬ﺒﺩﻻﻟﺔ ‪ ،n‬ﻨﻘﻭﻡ ﺒﺘﻌﻭﻴﺽ ‪ n‬ﺒﺎﻟﺭﺘﺒﺔ )‪ (n+1‬ﻓﻨﺠﺩ‪Un+1 = -2(n+1) + 5 = (-2n – :‬‬ ‫‪2) + 5 = -2n + 5‬‬‫ﺇﺫﻥ‪ Un+1 = -2n + (5-2) :‬ﻭﻤﻨﻪ‪Un+1 = -2n + 3 :‬‬ ‫‪ -‬ﺤﺴﺎﺏ ﺍﻟﻔﺭﻕ )‪(Un+1 – Un‬‬‫ﻟﺩﻴﻨﺎ‪Un+1- Un = (-2n + 3) – (-2n + 5) :‬‬‫ﻭﻤﻨﻪ‪Un+1 - Un = -2n + 3 –(-2n) – (+5):‬‬‫ﻭﻤﻨﻪ‪Un+1 – Un = -2n + 3+ 2n - 5 :‬‬‫ﺒﻌﺩ ﺍﻻﺨﺘﺯﺍل‪Un+1 – Un = 3-5 :‬‬ ‫ﺇﺫﻥ‪Un+1 – Un = -2 :‬‬‫ﻭﺒﻤﺎ ﺃﻥ ﺍﻟﻔﺭﻕ )‪ (Un+1 – Un‬ﻫﻭ ﻋﺩﺩ ﺴﺎﻟﺏ ﺘﻤﺎﻤﺎ ﺇﺫﻥ ‪:‬‬‫‪ Un+1 – Un < 0‬ﻤﻥ ﺃﺠل ﻜل ‪ n‬ﻤﻥ ‪ N‬ﻓﺈﻥ ﺍﻟﻤﺘﺘﺎﻟﻴﺔ )‪ (Un‬ﻤﺘﻨﺎﻗﺼﺔ ﺘﻤﺎﻤﺎ‪.‬‬ ‫‪Un = 2n (2‬‬ ‫‪-‬ﻜﺘﺎﺒﺔ ﺍﻟﺤﺩ ‪ Un+1‬ﺒﺩﻻﻟﺔ ‪n‬‬ ‫ﻟﺩﻴﻨﺎ‪Un+1 = 2(n+1) = 2n x 2 :‬‬‫‪Un+1 = 2 x 2n‬‬ ‫)ﺨﻭﺍﺹ ﺍﻟﻘﻭﻯ ﺍﻟﺼﺤﻴﺤﺔ(‬ ‫‪ -‬ﺤﺴﺎﺏ ﺍﻟﻔﺭﻕ‪(Un+1 – Un) :‬‬‫ﻟﺩﻴﻨﺎ‪Un+1 – Un = [2 x 2n] -[ 2n] :‬‬ ‫ﻤﻼﺤﻅﺔ ﺃﻥ ‪ 2n‬ﻫﻭ ﻋﺎﻤل ﻤﺸﺘﺭﻙ ﺇﺫﻥ‪:‬‬ ‫‪Un+1 – Un = 2n [2-1] = 2n [1] = 2n‬‬‫* ﺒﻤﺎ ﺃﻥ ﺍﻟﻌﺩﺩ ‪ 2n‬ﻤﻭﺠﺏ ﺘﻤﺎﻤﺎ ﺇﺫﻥ‪Un+1 – Un > 0 :‬‬ ‫ﻭﻤﻨﻪ )‪ (Un‬ﻤﺘﺯﺍﻴﺩﺓ ﺘﻤﺎﻤﺎ‪.‬‬ ‫‪Un‬‬ ‫=‬ ‫(‬ ‫‪1‬‬ ‫‪)n‬‬ ‫‪(4‬‬ ‫‪3‬‬ ‫‪ -‬ﻜﺘﺎﺒﺔ ﺍﻟﺤﺩ ‪ Un+1‬ﺒﺩﻻﻟﺔ ‪: n‬‬

‫‪Un+1‬‬ ‫=‬ ‫(‬ ‫‪1‬‬ ‫‪)n+1‬‬ ‫=‬ ‫(‬ ‫‪1‬‬ ‫‪)n‬‬ ‫‪x‬‬ ‫(‬ ‫‪1‬‬ ‫)‬ ‫ﻟﺩﻴﻨﺎ‪:‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫‪Un+1‬‬ ‫(=‬ ‫‪1‬‬ ‫)‬ ‫(‬ ‫‪1‬‬ ‫‪)n‬‬ ‫)ﺘﻁﺒﻴﻕ ﺨﻭﺍﺹ ﺍﻟﻘﻭﻯ(‬ ‫‪3‬‬ ‫‪3‬‬ ‫‪ -‬ﺤﺴﺎﺏ ﺍﻟﻔﺭﻕ‪(Un+1 – Un) :‬‬ ‫‪Un+1-‬‬ ‫‪Un‬‬ ‫(=‬ ‫‪1‬‬ ‫)‬ ‫(‬ ‫‪1‬‬ ‫‪)n‬‬ ‫–‬ ‫(‬ ‫‪1‬‬ ‫‪)n‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫ﺇﺫﻥ‪:‬‬ ‫ﻤﺸﺘﺭﻙ‬ ‫ﻋﺎﻤل‬ ‫ﻫﻭ‬ ‫(‬ ‫‪1‬‬ ‫‪)n‬‬ ‫ﺃﻥ‬ ‫ﻻﺤﻅ‬ ‫‪3‬‬‫‪Un+1‬‬ ‫–‬ ‫‪Un‬‬ ‫=‬ ‫(‬ ‫‪1‬‬ ‫‪)n‬‬ ‫[‬ ‫‪1‬‬ ‫]‪-1‬‬ ‫=‬ ‫(‬ ‫‪1‬‬ ‫)‬ ‫[‬ ‫‪1‬‬ ‫‬ ‫‪3‬‬ ‫]‬ ‫=‬ ‫(‬ ‫‪1‬‬ ‫‪)n‬‬ ‫[‬ ‫‪2‬‬ ‫]‬ ‫‪3‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫‪1‬‬ ‫‪)n‬‬ ‫(‬ ‫‪2‬‬ ‫)‬ ‫ﺇﺫﻥ ‪:‬‬ ‫‪Un+1 – Un =( 3‬‬ ‫‪3‬‬ ‫ﺘﻤﺎﻤﺎ‬ ‫ﻤﻭﺠﺏ‬ ‫(‬ ‫‪1‬‬ ‫‪)n‬‬ ‫ﺍﻟﻌﺩﺩ‬ ‫ﺃﻥ‬ ‫‪3‬‬ ‫ﺘﻤﺎﻤﺎ‬ ‫ﺴﺎﻟﺏ‬ ‫(‬ ‫‪2‬‬ ‫)‬ ‫ﺍﻟﻌﺩﺩ‬ ‫ﺃﻥ‬ ‫‪3‬‬ ‫ﺇﻥ ﺠﺩﺍﺅﻫﻤﺎ ﺴﺎﻟﺏ ﺘﻤﺎﻤﺎ ﺇﺫﻥ‪(Un+1 – Un) < 0 :‬‬ ‫ﻭﻤﻨﻪ )‪ (Un‬ﻤﺘﻨﺎﻗﺼﺔ ﺘﻤﺎﻤﺎ ﻋﻠﻰ ‪N‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ ‪: 5‬‬ ‫ﻫل ﺍﻟﻤﺘﺘﺎﻟﻴﺔ )‪ (Un‬ﺍﻟﻤﻌﺭﻓﺔ ﻋﻠﻰ )‪ (Un‬ﺤﺴﺎﺒﻴﺔ ؟‬ ‫‪(1)....... Un = 3n -7‬‬ ‫‪ -‬ﻜﺘﺎﺒﺔ ‪ Un+1‬ﺒﺩﻻﻟﺔ ‪n‬‬ ‫ﻟﺩﻴﻨﺎ ‪:‬‬ ‫‪Un+1 = 3 ( n+1) -7‬‬ ‫ﻭ ﻤﻨﻪ ‪Un+1 = 3n + 3 -7 :‬‬ ‫‪Un+1 = 3n -4‬‬ ‫ﺇﺫﻥ‬

‫‪ -‬ﺤﺴﺎﺏ ﺍﻟﻔﺭﻕ ‪(Un+1 – Un ) :‬‬‫) ‪Un+1 – Un = 3n -4- (3n -7‬‬ ‫‪= 3n – 4 – 3n + 7‬‬ ‫‪Un+1 – Un = 3‬‬ ‫ﺇﺫﻥ‬‫ﻭ ﻤﻨﻪ ) ‪ (Un‬ﻫﻲ ﻤﺘﺘﺎﻟﻴﺔ ﺤﺴﺎﺒﻴﺔ ﺃﺴﺎﺴﻬﺎ ‪r = 3‬‬‫‪(3) Un = n2‬‬ ‫‪ -‬ﻜﺘﺎﺒﺔ ‪ Un+1‬ﺒﺩﻻﻟﺔ ‪n‬‬ ‫ﻟﺩﻴﻨﺎ ‪Un+1 = (n+1)2 :‬‬ ‫‪= n2 +2 (n)(1) + 12‬‬‫= ‪Un+1‬‬ ‫‪n2 + 2(n)(1) + 12‬‬‫= ‪Un+1‬‬ ‫‪n2 + 2n) +1‬‬ ‫ﺇﺫﻥ‬ ‫‪ -‬ﺤﺴﺎﺏ ﺍﻟﻔﺭﻕ ) ‪(Un+1 – Un‬‬‫ﻟﺩﻴﻨﺎ )‪Un+1-Un = n2 + 2n + 1 - (n2‬‬ ‫‪Un+1-Un = 2n + 1‬‬ ‫ﺇﺫﻥ‬‫ﺇﻥ ﺍﻟﻔﺭﻕ ‪ Un+1- Un‬ﻟﻴﺱ ﻋﺩﺩﺍ ﺜﺎﺒﺘﺎ‪ ،‬ﺇﺫﻥ )‪ (Un‬ﻟﻴﺴﺕ ﻤﺘﺘﺎﻟﻴﺔ ﺤﺴﺎﺒﻴﺔ‬ ‫‪(4) U0 = - 3‬‬ ‫‪Un+1 =Un + n‬‬ ‫ﻟﺩﻴﻨﺎ ‪ (Un) :‬ﻤﻌﺭﻓﺔ ﺒﻌﻼﻗﺔ ﺘﺭﺍﺠﻌﻴﺔ‬ ‫ﺇﺫﻥ ‪Un+1-Un = n :‬‬‫ﺇﻥ ﺍﻟﻔﺭﻕ )‪ (Un+1-Un‬ﻟﻴﺱ ﻋﺩﺩﺍ ﺜﺎﺒﺘﺎ ﻭ ﻤﻨﻪ )‪ (Un‬ﻟﻴﺴﺕ ﻤﺘﺘﺎﻟﻴﺔ ﺤﺴﺎﺒﻴﺔ‬

‫‪(5) U0 = 1‬‬ ‫‪Un+1 = Un+ 2‬‬ ‫)‪ (Un‬ﻤﻌﺭﻓﺔ ﺒﻌﻼﻗﺔ ﺘﺭﺍﺠﻌﻴﺔ‬ ‫ﺇﺫﻥ ‪Un+1 – Un = 2 :‬‬‫ﻭ ﻤﻨﻪ ﺍﻟﻔﺭﻕ ﻫﻭ ﻋﺩﺩ ﺜﺎﺒﺕ ﺇﺫﻥ )‪ (Un‬ﻫﻲ ﻤﺘﺘﺎﻟﻴﺔ ﺤﺴﺎﺒﻴﺔ ﺃﺴﺎﺴﻬﺎ ‪r=2‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ ‪: 6‬‬‫‪ ( Un ) (1‬ﻤﺘﺘﺎﻟﻴﺔ ﺤﺴﺎﺒﻴﺔ ﻤﻌﺭﻓﺔ ﻋﻠﻰ ‪N‬‬ ‫ﻜﺘﺎﺒﺔ ﺍﻟﺤﺩ ﺍﻟﻌﺎﻡ ‪ Un‬ﺒﺩﻻﻟﺔ ‪n‬‬ ‫‪1) U0 = -1 ,r = 4‬‬ ‫‪Un = U0 + n r‬‬ ‫‪Un = -1 + 4 n‬‬‫)‪2‬‬ ‫‪U0‬‬ ‫=‬ ‫‪3‬‬ ‫‪,‬‬ ‫=‪r‬‬ ‫‪-‬‬ ‫‪5‬‬ ‫‪2‬‬‫‪Un = U0 + nr‬‬ ‫ﻟﺩﻴﻨﺎ ‪:‬‬‫=‪Un‬‬ ‫‪3‬‬ ‫‪-‬‬ ‫‪5‬‬ ‫‪n‬‬ ‫‪2‬‬‫‪ (Un) (2‬ﻤﻌﺭﻓﺔ ﻋﻠﻰ *‪ N‬ﺤﺩﻫﺎ ﺍﻷﻭل ﻫﻭ ‪U1‬‬ ‫ﻜﺘﺎﺒﺔ ‪ Un‬ﺒﺩﻻﻟﺔ ‪n‬‬ ‫‪1 ) U1 = 2 , r = -2‬‬ ‫ﻟﺩﻴﻨﺎ ‪Un = U1 + (n -1)r :‬‬ ‫ﻭ ﻤﻨﻪ‪Un = 2 + (n-1) (-2) :‬‬ ‫ﺇﺫﻥ‪Un = 2 – 2n + 2 :‬‬

‫ﻭ ﻤﻨﻪ‪Un= 4 – 2n :‬‬ ‫‪2 ) U1 = 3 , r = 2‬‬ ‫ﻟﺩﻴﻨﺎ‪Un = U1 + (n-1) r :‬‬ ‫ﻭ ﻤﻨﻪ‪Un = 3 + (n-1)+2 :‬‬‫ﺇﺫﻥ ‪ Un = 3 + 2n -2 :‬ﻭ ﻤﻨﻪ ‪Un = (3-2) + 2n‬‬‫ﺇﺫﻥ ‪U n = 1 + 2n :‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ ‪: 7‬‬ ‫ﻟﻨﺎ‪U5 = 9, U2 = 3 :‬‬ ‫ﺤﺴﺎﺏ ﺍﻷﺴﺎﺱ ‪:r‬‬‫ﻟﺩﻴﻨﺎ ﺍﻟﻌﻼﻗﺔ ﺒﻴﻥ ﻜل ﺤﺩﻴﻥ ﻤﻥ ﺤﺩﻭﺩ ﻤﺘﺘﺎﻟﻴﺔ ﺤﺴﺎﺒﻴﺔ‪:‬‬ ‫‪U5 = U2 + ( 5-2 ) r‬‬ ‫ﻭ ﻤﻨﻪ‪9 = 3 + 3 r :‬‬ ‫‪9–3=3r‬‬ ‫ﻭ ﻤﻨﻪ‪6 = 3 r :‬‬‫‪r‬‬ ‫‪6‬‬ ‫‪2‬‬ ‫ﺇﺫﻥ ‪:‬‬ ‫‪3‬‬ ‫ﻭﻤﻨﻪ ‪r =2‬‬ ‫‪ -2‬ﺤﺴﺎﺏ ﺍﻟﺤﺩ ﺍﻷﻭل ‪: U0‬‬‫‪Un = U0 + 2 r‬‬ ‫ﻟﻨﺎ ﻤﻥ ﻋﺒﺎﺭﺓ ﺍﻟﺤﺩ ﺍﻟﻌﺎﻡ ‪:‬‬ ‫ﺒﺈﻋﻁﺎﺀ ﺍﻟﻘﻴﻤﺔ ‪ 2‬ﻟـ ‪ n‬ﻨﺠﺩ ‪:‬‬ ‫‪U2 = U0 + 2 r‬‬ ‫ﺇﺫﻥ ‪:‬‬ ‫)‪3 = U0 + 2(2‬‬

‫‪U0 = 3 – 4‬‬ ‫ﻭ ﻤﻨﻪ‪:‬‬ ‫‪U0 = -1‬‬ ‫ﺇﺫﻥ‪:‬‬ ‫‪ -2‬ﻜﺘﺎﺒﺔ ‪ Un‬ﺒﺩﻻﻟﺔ ‪: n‬‬ ‫‪Un = U0 +n r‬‬ ‫‪Un = -1 + 2 n‬‬ ‫ﻭ ﻤﻨﻪ‪:‬‬ ‫‪ – 4‬ﺤﺴﺎﺏ ﺍﻟﻤﺠﻤﻭﻉ ‪:Sn‬‬ ‫‪Sn = U0 +…+Un‬‬ ‫@ ‪Sn n21>U0 U n‬‬‫‪Sn‬‬ ‫‪n 1‬‬ ‫‪>1‬‬ ‫‪1‬‬ ‫‪2n‬‬ ‫@‬ ‫‪2‬‬ ‫‪Sn‬‬ ‫‪n 1‬‬ ‫‪>2‬‬ ‫‪2‬‬ ‫‪n‬‬ ‫@‬ ‫‪2‬‬‫ﻭﺒﺎﺨﺘﺯﺍل ﺍﻟﻌﺩﺩ ‪ 2‬ﻨﺠﺩ ‪:‬‬ ‫‪Sn‬‬ ‫‪n 1‬‬ ‫‪>2 1n‬‬ ‫@‬ ‫‪2‬‬ ‫)‪Sn = (n+1) (-1+n) = (n-1)(n+1‬‬ ‫‪Sn = n2 – 1‬‬ ‫‪ -5‬ﺍﺴﺘﻨﺘﺎﺝ ‪S20,S10 :‬‬ ‫‪S10 = (10)2 -1 = 100 -1 = 99‬‬ ‫‪S20=(2 0)2 -1=400-1=399‬‬ ‫اﻟﺘﻤﺮﻳﻦ ‪: 8‬‬ ‫ﻟﺪﻳﻨﺎ ‪U3+U4+U5=33 .....(1) :‬‬ ‫ﻭ )‪(U3)2+(U4)2+(U5 )2=381 ...(2‬‬ ‫ﺒﻤﺎ ﺃﻥ ‪ U4‬ﻫﻭ ﺍﻟﻭﺴﻁ ﺍﻟﺤﺴﺎﺒﻲ‬

‫‪ U3+U5=2 U4‬ﺒﺎﻟﺘﻌﻭﻴﺽ ﻓﻲ ﺍﻟﻤﻌﺎﺩﻟﺔ )‪ (1‬ﻨﺠﺩ‪:‬‬ ‫ﻟﺩﻴﻨﺎ‬ ‫‪ U4 +2 U4 =33‬ﻭﻤﻨﻪ ‪ 3 U4=33 :‬ﺇﺫﻥ ‪:‬‬‫‪ U 4‬ﻨﻌﻭﺽ ﺒﻘﻴﻤﺔ ‪ U4‬ﻓﻲ )‪(1‬ﻭ)‪(2‬ﻨﺠﺩ ‪:‬‬ ‫‪33‬‬ ‫‪11‬‬ ‫‪3‬‬ ‫‪ U3+11+U5=33‬ﻭ ‪(U3)2+(11)2+(U5)2=381‬‬ ‫)‪U3 +U5=33-11 =22 ....(1‬‬ ‫وﻣﻨﻪ‬ ‫)‪(U3)2+(U5)2 =381-121 =260 ...(2‬‬ ‫ﻭﻟﺩﻴﻨﺎ ‪ U5=U4+r :‬و ‪ U3=U4 – r‬ﻭﺒﺎﻟﺘﻌﻭﻴﺽ ﻓﻲ)‪ (2‬ﻨﺠﺩ ‪:‬‬ ‫‪(U4-r)2 + (U4+r)2 = 260‬‬‫‪(U4)2-2rU4 + r2 + (U4)2 +2r(U4)+r2=260‬‬ ‫‪2(U4)2+2r2=260‬‬ ‫وﺏﻌﺪ اﻹﺧﺘﺰال ﻥﺠﺪ ‪:‬‬‫إذن ‪ 2 (121)+2r2=260‬وﻣﻨﻪ ‪2r2=260-242‬‬‫‪r2‬‬ ‫‪18‬‬ ‫‪9‬‬ ‫إذن ‪:‬‬ ‫وﻣﻨﻪ ‪2r2=18‬‬ ‫‪2‬‬ ‫وﺏﻤﺎ أن ‪ r ¢0‬ﻓﺈن ‪r 3 :‬‬‫وﻋﻠﻴﻪ ﻓﺈن ‪ U3 11(3) 14 :‬و ‪U 5 11  (3) 8‬‬ ‫اﻟﺘﻤﺮﻳﻦ ‪: 9‬‬ ‫) ‪ (Un‬ﻣﺘﺘﺎﻟﻴﺔ ﺣﺴﺎﺏﻴﺔ ﻣﻌﺮﻓﺔ ﻋﻠﻰ ‪ ، N‬ﺏﺤﻴﺚ ‪r=3‬‬ ‫‪U1 + U2 + U3 + U4 = 34‬‬ ‫و‬ ‫‪ -1‬ﺣﺴﺎب ‪:U0‬‬‫‪ -‬ﻥﻜﺘﺐ آﻞ ﺣﺪ ﻣﻦ ﺣﺪود اﻟﻤﺠﻤﻮع ﺏﺪﻻﻟﺔ اﻷﺱﺎس ‪ r‬و اﻟﺤﺪ اﻷول ‪U0‬‬ ‫‪Un = U0 + n r‬‬ ‫ﻟﻨﺎ ‪:‬‬ ‫و ﻣﻨﻪ‪ :‬ﻟﻤﺎ ‪ n = 1‬ﻥﺠﺪ‪:‬‬ ‫‪U1 = U0 + r‬‬ ‫ﻟﻤﺎ ‪ n = 2‬ﻥﺠﺪ‪:‬‬ ‫‪U2 = U0 + 2r‬‬ ‫ﻟﻤﺎ ‪ n = 3‬ﻥﺠﺪ‪:‬‬ ‫‪U3 = U0 + 3r‬‬ ‫ﻟﻤﺎ ‪ n = 4‬ﻥﺠﺪ‪:‬‬

‫‪U4 = U0 + 4r‬‬ ‫إذن ﻥﻌﻮض ﺏﻘﻴﻤﺔ آﻞ ﺣﺪ ﻓﻲ اﻟﻤﺠﻤﻮع ﻓﻨﺠﺪ‪:‬‬‫‪(U0 + r) + (U0 + 2 r) + (U0 + 3 r) + (U0 + 4 r) = 34‬ﺏﻤﺎ أن ‪r=3‬‬ ‫ﻓﻨﺠﺪ ‪:‬‬‫‪(U0 + 3) + (U0 + 2 ) + (U0 + 9 ) + (U0 + 12 ) = 34‬‬ ‫‪4 U0 + 30 = 34‬‬ ‫و ﻣﻨﻪ‪:‬‬ ‫و ﻣﻨﻪ‪:‬‬ ‫‪4U0 = 34 – 30‬‬ ‫إذن ‪:‬‬ ‫‪4U0 = 4‬‬ ‫‪U0‬‬ ‫‪4‬‬ ‫‪4‬‬ ‫و ﻣﻨﻪ ‪U0 = 1 :‬‬ ‫‪ -2‬آﺘﺎﺏﺔ ‪ Un‬ﺏﺪﻻﻟﺔ ‪: n‬‬ ‫‪Un = U0 + nr‬‬ ‫‪Un = 1 + 3n‬‬ ‫و ﻣﻨﻪ‪:‬‬ ‫‪ -3‬ﺣﺴﺎب اﻟﻤﺠﻤﻮع‪:‬‬ ‫‪Sn = U0 + ……+ Un‬‬ ‫@ ‪Sn n21>U0 U n‬‬‫‪Sn‬‬ ‫‪n 1‬‬ ‫‪>2‬‬ ‫‪3n‬‬ ‫@‬ ‫‪2‬‬ ‫ﺣﺴﺎب اﻟﻤﺠﻤﻮع ‪S cn :‬‬ ‫‪S cn = U1+……+Un‬‬ ‫ﺣﺴﺎب ‪:U1‬‬ ‫إذن‪:‬‬ ‫‪U1 = U0 + r = 1 + 3 = 4‬‬ ‫‪U1=4‬‬

‫ﻟﺪﻳﻨﺎ‪:‬‬‫‪n‬‬‫‪> @ > @Scn‬‬ ‫‪n‬‬‫‪2‬‬ ‫‪U1 U n‬‬ ‫=‬ ‫‪2‬‬ ‫) ‪4  (1 3n‬‬ ‫‪S cn‬‬ ‫‪n‬‬ ‫@)‪>(53n‬‬ ‫‪2‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ ‪:10‬‬ ‫‪V0 = 300‬‬ ‫‪(1‬ﺣﺴﺎب ﻋﺪد اﻟﻌﻤﺎل ﻋﺎم ‪: 2001‬‬ ‫أي ﺣﺴﺎب ‪ V1‬و ﻣﻨﻪ‪:‬‬ ‫‪V1 = V0 + 40‬‬ ‫‪V1 = 300 + 40 = 340‬‬ ‫ﺣﺴﺎب ﻋﺪد اﻟﻤﺎل ﻋﺎم ‪: 2002‬‬ ‫أي ﺣﺴﺎب ‪:V2‬‬ ‫‪V2 = V1 + 40‬‬ ‫‪V2 = 380 + 40 = 420‬‬ ‫‪ (2‬إﻳﺠﺎد اﻟﻌﻼﻗﺔ ﺏﻴﻦ ‪ Vn + 1‬و ‪:Vn‬‬ ‫ﻟﺪﻳﻨﺎ اﻻﺱﺘﻨﺘﺎج‪:‬‬ ‫‪ :Vn + 1‬ﻋﺪد اﻟﻌﻤﺎل ﺏﻌﺪ )‪( n+1‬ﺱﻨﺔ‬ ‫‪ :Vn‬ﻋﺪد اﻟﻌﻤﺎل ﺏﻌﺪ ‪ n‬ﺱﻨﺔ‬ ‫‪Vn+1=Vn +40‬‬ ‫إدن ‪:‬‬‫‪ (3‬ﻣﻨﻪ )‪ (Vn‬هﻲ م‪.‬ح أﺱﺎﺱﻬﺎ ‪ r = 40‬و ﺣﺪهﺎ اﻷول هﻮ ‪V0 = 300‬‬ ‫‪Vn + 1 = Vn + 40‬‬ ‫‪(4‬آﺘﺎﺏﺔ ‪ Vn‬ﺏﺪﻻﻟﺔ ‪:n‬‬ ‫ﻟﺪﻳﻨﺎ ‪Vn = V0 + nr‬‬ ‫أي‪Vn = 300 + 40n :‬‬ ‫‪ (5‬ﺣﺴﺎب اﻟﻤﺠﻤﻮع ‪Sn‬‬ ‫‪Sn = V0 +V1 +…+V n‬‬

‫‪> @Sn‬‬ ‫‪n 1‬‬ ‫‪2‬‬ ‫‪V0 Vn‬‬‫‪Sn‬‬ ‫‪n 1‬‬ ‫‪>300(300‬‬ ‫‪40n‬‬ ‫@)‬ ‫‪2‬‬‫‪Sn‬‬ ‫‪n 1‬‬ ‫‪>2(300‬‬ ‫‪ 40‬‬ ‫@)‪n‬‬ ‫‪2‬‬ ‫ﺒﻌﺩ ﺇﺨﺘﺯﺍل ﺍﻟﻌﺩﺩ‪ 2‬ﺒﺴﻁﺎ ﻭﻤﻘﺎﻤﺎ ﻨﺠﺩ‬ ‫@)‪S n (n  1)>(30040n‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ ‪: 11‬‬ ‫ﻟﻨﺎ ‪ U0 = 3000‬اﻟﻤﺒﻠﻎ اﻟﻤﻮدع ﻋﺎم ‪2007‬‬ ‫اﻟﻤﺒﻠﻎ اﻟﻤﺤﺼﻞ ﻋﺎم ‪:2008‬‬ ‫‪U1 = U0 +(0.05)3000‬‬ ‫‪U1 = U0 +3150‬‬ ‫اﻟﻤﺒﻠﻎ اﻟﻤﺤﺼﻞ ﻋﺎم ‪:2009‬‬ ‫‪U2 = U1 +( 0.05)3000‬‬ ‫‪U2 = 3150 + 150 = 3300 DA‬‬ ‫‪ -2‬اﻟﻌﻼﻗﺔ ﺏﻴﻦ ‪ Un +1‬و ‪: Un‬‬ ‫‪ :Un +1‬اﻟﻤﺒﻠﻎ اﻟﻤﺤﺼﻞ ﺏﻌﺪ ‪ n+1‬ﺱﻨﺔ‬ ‫‪ : Un‬اﻟﻤﺒﻠﻎ اﻟﻤﺤﺼﻞ ﺏﻌﺪ ‪ n‬ﺱﻨﺔ‬ ‫إذن ‪:‬‬ ‫‪Un +1 = Un + (0.05) 300‬‬ ‫‪Un +1 = Un + 150‬‬‫و ﻣﻨﻪ )‪ (Un‬ﻣﺘﺘﺎﻟﻴﺔ ﺣﺴﺎﺏﻴﺔ أﺱﺎﺱﻬﺎ ‪ r = 150‬و ﺣﺪهﺎ اﻷول هﻮ ‪U0 = 2000‬‬ ‫‪Un = U0 + n r‬‬ ‫و ﻣﻨﻪ‪:‬‬ ‫‪Un = 2000 + 150n‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ ‪:12‬‬

‫) ‪ (Un‬ﻣﺘﺘﺎﻟﻴﺔ ﻣﻌﺮﻓﺔ ﻋﻠﻰ ‪N‬‬ ‫‪ ،U0 = 1‬ﻟﻨﺎ اﻟﻤﺠﻤﻮع‪:‬‬ ‫‪S = 1 + 1 + 21 + …. + 2001‬‬ ‫‪ -1‬ﺣﺴﺎب اﻷﺱﺎس ‪:r‬‬ ‫‪r = 11 – 1 = 10‬‬ ‫‪ -2‬آﺘﺎﺏﺔ ‪ Un‬ﺏﺪﻻﻟﺔ ‪: n‬‬ ‫‪Un = U0 + n r‬‬ ‫‪Un = 1 + n 10‬‬ ‫‪ -3‬ﻋﻴﻦ ‪ n‬ﺏﺤﻴﺚ‪Un = 201 :‬‬ ‫ﻟﻨﺎ‪Un = 201 :‬‬ ‫و ﻣﻨﻪ‪1 + 10n = 201 :‬‬ ‫ﻟﺪﻳﻨﺎ ﻣﻌﺎدﻟﺔ ﻣﻦ اﻟﺪرﺝﺔ اﻷوﻟﻰ ذات اﻟﻤﺠﻬﻮل اﻟﻄﺒﻴﻌﻲ ‪ n‬و ﻣﻨﻪ‪:‬‬ ‫‪10n = 201 – 1‬‬ ‫إذن‪10 n = 200 :‬‬ ‫‪n‬‬ ‫‪200‬‬ ‫‪20‬‬ ‫أي‪:‬‬ ‫‪10‬‬ ‫و ﻣﻨﻪ ‪n = 20 :‬‬ ‫‪Sn = U0 + ……+ Un‬‬ ‫إذن‪U20 = 201 :‬‬ ‫‪ -4‬ﺣﺴﺎب ‪:Sn‬‬ ‫@ ‪Sn n21>U 0 U n‬‬ ‫ﻟﺪﻳﻨﺎ‬ ‫وﻣﻨﻪ‬‫‪Sn‬‬ ‫‪n 1‬‬ ‫‪>1110‬‬ ‫@‪n‬‬ ‫‪2‬‬ ‫‪> @S n‬‬‫‪n 1‬‬ ‫>‬ ‫=‬ ‫‪n 1‬‬ ‫‪2‬‬ ‫@ ‪2 10 n‬‬ ‫‪2‬‬ ‫)‪2 (15n‬‬ ‫وﺏﺎﺧﺘﺰال اﻟﻌﺪد ‪ 2‬ﺏﺴﻄﺎ وﻣﻘﺎﻣﺎ ﻥﺠﺪ ‪:‬‬

‫@)‪S n (n  1)>(15n‬‬ ‫ﻥﺠﺪ ﺏﻌﺪ اﻟﻨﺸﺮ‪:‬‬ ‫‪Sn = n + 5n2 + 1 + 5n‬‬ ‫‪Sn = 5n2 + 6n + 1‬‬ ‫‪Sn = 105‬‬ ‫‪ -3‬هﻞ ﻳﻮﺝﺪ ‪ n‬ﺏﺤﻴﺚ ‪:‬‬ ‫ﻟﺘﻜﻦ اﻟﻤﻌﺎدﻟﺔ ذات اﻟﻤﺠﻬﻮل اﻟﻄﺒﻴﻌﻲ ‪: n‬‬ ‫‪5n2 + 6n + 1 = 105‬‬ ‫‪5n2 + 6n – 104 = 0‬‬ ‫‪a = 5, b = 6, c = - 104‬‬ ‫ﺣﺴﺎب اﻟﻤﻤﻴﺰ ‪:‬‬ ‫)‪∆ = b2 – 4 a c = (6)2 – 4(5)(-104‬‬ ‫)‪= 36 + (20 )(104‬‬ ‫‪∆ = 36 + 2080 = 216‬‬ ‫‪' 46‬‬ ‫إذن‪:‬‬ ‫و ﻣﻨﻪ ﻟﻠﻤﻌﺎدﻟﺔ ﺣﻠﻴﻦ ﻣﺘﻤﺎﻳﺰﻳﻦ ‪:‬‬ ‫‪n1‬‬ ‫' ‪b‬‬ ‫‪ 6  46‬‬ ‫‪40‬‬ ‫‪4‬‬ ‫‪2a‬‬ ‫‪2u5‬‬ ‫‪10‬‬ ‫‪n1 = 4  N‬‬‫‪n1‬‬ ‫' ‪b‬‬ ‫‪ 6  46‬‬ ‫‪52‬‬ ‫‬ ‫‪N‬‬ ‫‪2a‬‬ ‫‪2u5‬‬ ‫‪10‬‬ ‫إذن اﻟﻌﺪد اﻟﻤﻄﻠﻮب هﻮ ‪n =4‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ ‪: 13‬‬ ‫هﻞ ) ‪ (Un‬ﻣﺘﺘﺎﻟﻴﺔ هﻨﺪﺱﻴﺔ؟‬ ‫‪1) Un = 3 . (2)n‬‬ ‫‪ -‬إﻳﺠﺎد اﻟﺤﺪ ‪: Un +1‬‬ ‫‪Un +1 = 3 . (2)n +1‬‬ ‫ﻟﻨﺎ ‪:‬‬ ‫و ﺣﺴﺐ ﺧﻮاص اﻟﻘﻮى‪:‬‬ ‫‪Un +1 = 3 . (2)n .(2) = (2) (3) (2)n‬‬