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amaths
發問:
Q1 2^2+4^2+6^2+...+=(2/3)n(n+1)(2n+1) 請寫清楚步驟及答案！
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Let P ( n ) be the proposition “22+42+62+...+(2n)2=(2/3)n(n+1)(2 n+1)”. When n = 1, L.H.S. = (2)2 = 4 R.H.S. = (2/3)(1)(1+1)(2 +1) = 4 = L.H.S. So P ( 1 ) is true. Assume P ( k ) is true for some positive integers k, i.e. 22+42+62+...+(2k)2=(2/3)k(k+1)(2 k+1) When n = k + 1, L H.S. = 22+42+62+...+(2k)2+(2k+2)2 =(2k/3)(k+1)(2k+1)+(2k+2)2 =(2/3)[k(k+1)(2k+1)+3(2k+2)2/2] =(2/3)[k(k+1)(2k+1)+6(k+1)2] =(2/3)(k+1)(2k2+k+6k+6) =(2/3)(k+1)(2k2+7k+6) =(2/3)(k+1)(k+2)(2k+3) R.H.S. = (2/3)(k+1)(k+1+1)(2k+2+1) =(2/3)(k+1)(k+2)(2k+3) =L.H.S. So P ( k + 1 ) is true. By the principle of mathematical induction, P ( n ) is true for all positive integers n.
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