關(guān)于 1^2 + 2^2 + …… + n^2 = n×(n+1)×(2n+1)/6 的證明：
(n+1)^3 = n^3 + 3n^2 + 3n + 1
所以
(n+1)^3 - n^3 = 3n^2 + 3n + 1
n^3 - (n-1)^3 = 3(n-1)^2 + 3(n-1) + 1
(n-1)^3 - (n-2)^3 = 3(n-2)^2 + 3(n-2) + 1
.
3^3 - 2^3 = 3*2^2 + 3*2 + 1
2^3 - 1^3 = 3*1^2 + 3*1 +1
把以上n個(gè)等式的兩邊分別相加得到
(n+1)^3-1^3 =
3×(1^2+2^2+3^2+...+n^2) + 3×(1+2+3+……+n) + n個(gè)1的和
(n+1)^3-1 = 3×(1^2+2^2+...+n^2) + 3×n×(n+1)/2 + n
所以
3(1^2+2^2+.+n^3)
= n^3 + 3n^2 + 3n - 3n(n+1)/2 - n
= n(n^2+3n+2) - 3n(n+1)/2
= n(n+1)(n+2)-3n(n+1)/2
= n(n+1)(2n+1)/2
最后
1^2+2^2+.+n^2 = n(n+1)(2n+1)/6.