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標題:
Mathematical Induction(1)
發問:
請幫我計一計以下這條數： 1a) Prove by mathematical induction that for all positive integers n, 1?3^2+2?5^2+3?7^2+...+n(2n+1)^2=1/6n(n+1)(6n^2+14n+7). b) Hence, or otherwise, find the sum of 11?23^2+12?25^2+13?27^2+...+20?41^2 更新: (PS?means *)
最佳解答:
其他解答:
1a) Prove by mathematical induction that for all positive integers n, 1?3^2+2?5^2+3?7^2+...+n(2n+1)^2=1/6n(n+1)(6n^2+14n+7). When n = 1, L.H.S. = 1 * ( 2 * 1 + 1 )^2 = 1 * 3^2 = 1 * 9 = 9 R.H.S. = (1/6) * 1 * (1+1) * (6*1^2 + 14*1 + 7 ) = (1/6) * 1 * 2 * (6 + 14 + 7) = (1/6) * 1 * 2 * 27 = 9 = L.H.S. So that the statement is true for n = 1. Assume that the statement is true for n = k. That is 1 * 3^2 + 2 * 5^2 + 3 * 7^2 + ... + k * (2*k+1)^2 = (1/6)* k * (k+1) * (6*k^2+14*k+7) Now to prove that n = k+1 is ture. That means 1 * 3^2 + 2 * 5^2 + 3 * 7^2 + ... + k * (2k+1)^2 + (k+1) * [2*(k+1)+1]^2 = (1/6) * (k+1) *[(k+1)+1] * [6 * (k+1)^2 + 14 * (k+1) + 7] L.H.S. = 1 * 3^2 + 2 * 5^2 + 3 * 7^2 + ... + k * (2*k+1)^2 + (k+1) * [2*(k+1)+1]^2 = (1/6) * k* (k+1) * (6*k^2+14*k+7) + (k+1) * [2*(k+1)+1]^2 = (1/6) * k* (k+1) * (6*k^2+14*k+7) + (k+1) * (2*k + 3)^2 = (1/6) * k* (k+1) * (6*k^2+14*k+7) + (k+1) * (4*k^2 + 12*k + 9) = (1/6) * (k+1) * [k * (6*k^2+14*k+7) + 6 * (4*k^2 + 12*k + 9)] = (1/6) * (k+1) * (6 * k^3 + 14 * k^2 + 7*k + 24 * k^2 + 72 * k + 54) = (1/6) * (k+1) * (6 * k^3 + 38 * k^2 + 79 * k + 54) = (1/6) * (k+1) * (k+2) * (6 * k^2 + 26 * k + 27) R.H.S. = (1/6) * (k+1) *[(k+1)+1] * [6 * (k+1)^2 + 14 * (k+1) + 7] = (1/6) * (k+1) * (k+2) * [6 * (k^2 + 2 * k + 1) + 14 * k + 14 + 7] = (1/6) * (k+1) * (k+2) * (6 * k^2 + 12 * k + 6 + 14 * k + 21) = (1/6) * (k+1) * (k+2) * (6 * k^2 + 26 * k + 27) = L.H.S. So that it is true for n = k+1 Therefore, the statment 1?3^2+2?5^2+3?7^2+...+n(2n+1)^2=1/6n(n+1)(6n^2+14n+7) is true for all positive integers n. b) Hence, or otherwise, find the sum of 11?23^2+12?25^2+13?27^2+...+20?41^2 = (1?3^2+2?5^2+3?7^2+ ... + 20?41^2) - (1?3^2+2?5^2+3?7^2+ ... + 10?21^2) = (1/6)?20?(20+1)?(6?20^2 + 14?20 + 7) - (1/6)?10?(10+1)?(6?10^2 + 14?10 + 7) = (1/6)?20?21?(6?400 + 280 + 7) - (1/6)?10?11?(6?100 + 140 + 7) = (1/6)?20?21?2687 - (1/6)?10?11? 747 = 188090 - 13695 = 174395
Mathematical Induction(1)
發問:
請幫我計一計以下這條數： 1a) Prove by mathematical induction that for all positive integers n, 1?3^2+2?5^2+3?7^2+...+n(2n+1)^2=1/6n(n+1)(6n^2+14n+7). b) Hence, or otherwise, find the sum of 11?23^2+12?25^2+13?27^2+...+20?41^2 更新: (PS?means *)
最佳解答:
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(a) For n = 1, LHS = 1(2 + 1)^2 = 9 RHS = (1 / 6)(1)(1 + 1)(6 + 14 + 7) = (1 / 6)(2)(27) = 9 = LHS. Thus it is true for n = 1. Assume that it is true for n = k, i.e. 1?3^2+2?5^2+3?7^2+...+k(2k+1)^2=(1/6)k(k+1)(6k^2+14k+7) For n = k + 1, LHS = 1?3^2+2?5^2+3?7^2+...+k(2k+1)^2 + (k + 1)[2(k + 1) + 1]^2 =(1/6)k(k + 1)(6k^2 + 14k +7) + (k + 1)[2k + 3]^2 = (1 / 6)(k + 1)[k(6k^2 + 14k +7) + 6(2k + 3)^2] = (1 / 6)(k + 1)[6k^3 + 14k^2 + 7k + 24k^2 + 72k + 54] = (1 / 6)(k + 1)[6k^3 + 38k^2 + 79k + 54] = (1 / 6)(k + 1)[6k^3 + 12k^2 + 26k^2 + 52k + 27k + 54] = (1 / 6)(k + 1)[6k^2(k + 2) + 26k(k + 2) + 27(k + 2)] = (1 / 6)(k + 1)(k + 2)(6k^2 + 26k + 27) = (1 / 6)(k + 1)(k + 2)(6k^2 + 12k + 6 + 14k + 14 + 7) = (1 / 6)(k + 1)(k + 2)[6(k + 1)^2 + 14(k + 1) + 7] = RHS. Thus it is true for n = k + 1. By the principle of mathematical induction, it is true for all positive integers n. (b) 11 * 23^2 + 12 * 25^2 + ... + 20 * 41^2 = 1 * 3^2 + 2 * 5^2 + ... + 10 * 21^2 + 11 * 23^2 + 12 * 25^2 + ... + 20 * 41^2 - (1 * 3^2 + 2 * 5^2 + ... + 10 * 21^2) = (1 / 6)(20)(20 + 1)[6(20^2) + 14(20) + 7] - (1 / 6)(10)(10 + 1)[6(10^2) + 14(10) + 7] = (1 / 6)(20)(21)(2687) - (1 / 6)(10)(11)(747) = 188090 - 13695 = 174395.其他解答:
1a) Prove by mathematical induction that for all positive integers n, 1?3^2+2?5^2+3?7^2+...+n(2n+1)^2=1/6n(n+1)(6n^2+14n+7). When n = 1, L.H.S. = 1 * ( 2 * 1 + 1 )^2 = 1 * 3^2 = 1 * 9 = 9 R.H.S. = (1/6) * 1 * (1+1) * (6*1^2 + 14*1 + 7 ) = (1/6) * 1 * 2 * (6 + 14 + 7) = (1/6) * 1 * 2 * 27 = 9 = L.H.S. So that the statement is true for n = 1. Assume that the statement is true for n = k. That is 1 * 3^2 + 2 * 5^2 + 3 * 7^2 + ... + k * (2*k+1)^2 = (1/6)* k * (k+1) * (6*k^2+14*k+7) Now to prove that n = k+1 is ture. That means 1 * 3^2 + 2 * 5^2 + 3 * 7^2 + ... + k * (2k+1)^2 + (k+1) * [2*(k+1)+1]^2 = (1/6) * (k+1) *[(k+1)+1] * [6 * (k+1)^2 + 14 * (k+1) + 7] L.H.S. = 1 * 3^2 + 2 * 5^2 + 3 * 7^2 + ... + k * (2*k+1)^2 + (k+1) * [2*(k+1)+1]^2 = (1/6) * k* (k+1) * (6*k^2+14*k+7) + (k+1) * [2*(k+1)+1]^2 = (1/6) * k* (k+1) * (6*k^2+14*k+7) + (k+1) * (2*k + 3)^2 = (1/6) * k* (k+1) * (6*k^2+14*k+7) + (k+1) * (4*k^2 + 12*k + 9) = (1/6) * (k+1) * [k * (6*k^2+14*k+7) + 6 * (4*k^2 + 12*k + 9)] = (1/6) * (k+1) * (6 * k^3 + 14 * k^2 + 7*k + 24 * k^2 + 72 * k + 54) = (1/6) * (k+1) * (6 * k^3 + 38 * k^2 + 79 * k + 54) = (1/6) * (k+1) * (k+2) * (6 * k^2 + 26 * k + 27) R.H.S. = (1/6) * (k+1) *[(k+1)+1] * [6 * (k+1)^2 + 14 * (k+1) + 7] = (1/6) * (k+1) * (k+2) * [6 * (k^2 + 2 * k + 1) + 14 * k + 14 + 7] = (1/6) * (k+1) * (k+2) * (6 * k^2 + 12 * k + 6 + 14 * k + 21) = (1/6) * (k+1) * (k+2) * (6 * k^2 + 26 * k + 27) = L.H.S. So that it is true for n = k+1 Therefore, the statment 1?3^2+2?5^2+3?7^2+...+n(2n+1)^2=1/6n(n+1)(6n^2+14n+7) is true for all positive integers n. b) Hence, or otherwise, find the sum of 11?23^2+12?25^2+13?27^2+...+20?41^2 = (1?3^2+2?5^2+3?7^2+ ... + 20?41^2) - (1?3^2+2?5^2+3?7^2+ ... + 10?21^2) = (1/6)?20?(20+1)?(6?20^2 + 14?20 + 7) - (1/6)?10?(10+1)?(6?10^2 + 14?10 + 7) = (1/6)?20?21?(6?400 + 280 + 7) - (1/6)?10?11?(6?100 + 140 + 7) = (1/6)?20?21?2687 - (1/6)?10?11? 747 = 188090 - 13695 = 174395
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