標題:
FORM 4 MI
發問:
1^2 + 2^2 + 3^2 + ... + n^2 = 1/6 n (n+1)(2n+1) using the result , find the sum of a). 2^2 + 4^2 + 6^2 + ...+ (2m)^2 b) . 1^2 + 3^2 + 5^2 + ...+ (2m - 1) ^2
最佳解答:
12 + 22 + 32 + ... + n2 = n(n+1)(2n+1)/6 using the result , find the sum of a). 22 + 42 + 62 + ...+ (2m)2 22 + 42 + 62 + ...+ (2m)2 = 22(12 + 22 + 32 + ... + m2) = 22m(m+1)(2m+1)/6 = 2m(m+1)(2m+1)/3 b) . 12 + 32 + 52 + ...+ (2m - 1)2 [12 + 32 + 52 + ...+ (2m - 1)2] + [22 + 42 + 62 + ...+ (2m)2] = 12 + 22 + 32 + ... + (2m-1)2 + 2m2 [12 + 32 + 52 + ...+ (2m - 1)2] = [12 + 22 + 32 + ... + (2m-1)2 + 2m2] - [22 + 42 + 62 + ...+ (2m)2] = (2m)(2m+1)(4m+1)/6 - 2m(m+1)(2m+1)/3 = m(2m+1)(4m+1)/3 - 2m(m+1)(2m+1)/3 = m[(2m+1)(4m+1) - 2(m+1)(2m+1)]/3 = m[8m2+ 6m+1 – 4m2 – 6m - 2]/3 = m[4m2 - 1]/3 = m(2m – 1)(2m + 1)/3
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