標題:
M.I. q
a) Prove, by M.I. ,that 3/4+5/36+7/144+...+2n+1/n^2(n+1)^2=n(n+2)/(n+1)^2 for all positive integers n. b) Hence, find the value of 13/1764+15/3136+17/5184+...+39/144400 .
最佳解答:
a) Let P(n) be 3/4+5/36+7/144+...+(2n+1)/n2(n+1)2=n(n+2)/(n+1)2 for all positive integers n. When n = 1, Left hand side (L.H.S.) = [2(1)+1]/[12(1+1)2] = (2+1)(4) = 3/4 Right hand side (R.H.S.) = 1(1+2)/(1+1)2 = 3/4 As L.H.S. = R.H.S., so P(1) is true. Assume P(k) is true, i.e. 3/4+5/36+7/144+...+(2k+1)/k2(k+1)2=k(k+2)/(k+1)2 for any positive integer k. When n = k+1, Left hand side (L.H.S.) = 3/4+5/36+7/144+...+(2k+1)/k2(k+1)2 + [2(k+1)+1]/[(k+1)2((k+1)+1)2] = k(k+2)/(k+1)2 + (2k+3)/[(k+1)2(k+2)2] ............ by assumsion of P(k) = k(k+2)^3/[(k+1)2(k+2)2] + (2k+3)/[(k+1)2(k+2)2] = 【k(k+2)^3 + 2k + 3】 / [(k+1)2(k+2)2] = 【k^4 + 6k^3 + 12k2 + 8k + 2k + 3】 / [(k+1)2(k+2)2] = 【k^4 + 6k^3 + 12k2 + 10k + 3】 / [(k+1)2(k+2)2] = 【(k^4 + 4k^3 + 6k2 + 4k + 1) + 2k^3 + 6k2 + 6k + 2】 / [(k+1)2(k+2)2] = 【(k+1)^4 + 2(k^3 + 3k2 + 3k + 1)】 / [(k+1)2(k+2)2] = 【(k+1)^4 + 2(k+1)^3】 / [(k+1)2(k+2)2] = (k+1)^3(k+1+2) / [(k+1)2(k+2)2] = (k+1)^3(k+3) / [(k+1)2(k+2)2] = (k+1)(k+3) / (k+2)2 = (k+1)[(k+1)+2] / [(k+1)+1]2 = R.H.S. So P(k+1) is true. By Mathematical induction, 3/4+5/36+7/144+...+2n+1/n2(n+1)2=n(n+2)/(n+1)2 is true for all positive integers n. ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ b) 13/1764+15/3136+17/5184+...+39/144400 = (2*6+1)/62(6+1)2 + (2*7+1)/72(7+1)2 + (2*8+1)/82(8+1)2 + ... + (2*19+1)/192(19+1)2 = 【(2*1+1)/12(1+1)2 + (2*2+1)/22(2+1)2 + (2*3+1)/32(3+1)2 + ... + (2*19+1)/192(19+1)2】 - 【(2*1+1)/12(1+1)2 + (2*2+1)/22(2+1)2 + (2*3+1)/32(3+1)2 + ... + (2*5+1)/52(5+1)2】 = 19(19+2)/(19+1)2 - 5(5+2)/(5+1)2 = 19(21)/202 - 5(7)/62 = 399/400 - 35/36 = (14364 - 14000)/14400 = 91/3600 2006-11-18 23:16:32 補充： b part 其實應用了 part a mathematical induction 的結果，只是要先看透這個算式是 (2n+1)/n2(n+1)2 這個 pattern 的第 6 項至第 19 項。只要將 【1-19 項的結果】 - 【1-5 項的結果】就可得出答案。
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M.I. q
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發問:a) Prove, by M.I. ,that 3/4+5/36+7/144+...+2n+1/n^2(n+1)^2=n(n+2)/(n+1)^2 for all positive integers n. b) Hence, find the value of 13/1764+15/3136+17/5184+...+39/144400 .
最佳解答:
a) Let P(n) be 3/4+5/36+7/144+...+(2n+1)/n2(n+1)2=n(n+2)/(n+1)2 for all positive integers n. When n = 1, Left hand side (L.H.S.) = [2(1)+1]/[12(1+1)2] = (2+1)(4) = 3/4 Right hand side (R.H.S.) = 1(1+2)/(1+1)2 = 3/4 As L.H.S. = R.H.S., so P(1) is true. Assume P(k) is true, i.e. 3/4+5/36+7/144+...+(2k+1)/k2(k+1)2=k(k+2)/(k+1)2 for any positive integer k. When n = k+1, Left hand side (L.H.S.) = 3/4+5/36+7/144+...+(2k+1)/k2(k+1)2 + [2(k+1)+1]/[(k+1)2((k+1)+1)2] = k(k+2)/(k+1)2 + (2k+3)/[(k+1)2(k+2)2] ............ by assumsion of P(k) = k(k+2)^3/[(k+1)2(k+2)2] + (2k+3)/[(k+1)2(k+2)2] = 【k(k+2)^3 + 2k + 3】 / [(k+1)2(k+2)2] = 【k^4 + 6k^3 + 12k2 + 8k + 2k + 3】 / [(k+1)2(k+2)2] = 【k^4 + 6k^3 + 12k2 + 10k + 3】 / [(k+1)2(k+2)2] = 【(k^4 + 4k^3 + 6k2 + 4k + 1) + 2k^3 + 6k2 + 6k + 2】 / [(k+1)2(k+2)2] = 【(k+1)^4 + 2(k^3 + 3k2 + 3k + 1)】 / [(k+1)2(k+2)2] = 【(k+1)^4 + 2(k+1)^3】 / [(k+1)2(k+2)2] = (k+1)^3(k+1+2) / [(k+1)2(k+2)2] = (k+1)^3(k+3) / [(k+1)2(k+2)2] = (k+1)(k+3) / (k+2)2 = (k+1)[(k+1)+2] / [(k+1)+1]2 = R.H.S. So P(k+1) is true. By Mathematical induction, 3/4+5/36+7/144+...+2n+1/n2(n+1)2=n(n+2)/(n+1)2 is true for all positive integers n. ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ b) 13/1764+15/3136+17/5184+...+39/144400 = (2*6+1)/62(6+1)2 + (2*7+1)/72(7+1)2 + (2*8+1)/82(8+1)2 + ... + (2*19+1)/192(19+1)2 = 【(2*1+1)/12(1+1)2 + (2*2+1)/22(2+1)2 + (2*3+1)/32(3+1)2 + ... + (2*19+1)/192(19+1)2】 - 【(2*1+1)/12(1+1)2 + (2*2+1)/22(2+1)2 + (2*3+1)/32(3+1)2 + ... + (2*5+1)/52(5+1)2】 = 19(19+2)/(19+1)2 - 5(5+2)/(5+1)2 = 19(21)/202 - 5(7)/62 = 399/400 - 35/36 = (14364 - 14000)/14400 = 91/3600 2006-11-18 23:16:32 補充： b part 其實應用了 part a mathematical induction 的結果，只是要先看透這個算式是 (2n+1)/n2(n+1)2 這個 pattern 的第 6 項至第 19 項。只要將 【1-19 項的結果】 - 【1-5 項的結果】就可得出答案。
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