求1²+2²+3²+4²+······+n²=?

求1²+2²+3²+4²+······+n²=?
求1²+2²+3²+4²+······+n²=?

求1²+2²+3²+4²+······+n²=?
这是个公式
设S=1^2+2^2+.+n^2
(n+1)^3-n^3 = 3n^2+3n+1
n^3-(n-1)^3 = 3(n-1)^2+3(n-1)+1
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2^3-1^3 = 3*1^2+3*1+1
把上面n个式子相加得：(n+1)^3-1 = 3* [1^2+2^2+...+n^2] +3*[1+2+.+n] +n
所以S= (1/3)*[(n+1)^3-1-n-(1/2)*n(n+1)] = (1/6)n(n+1)(2n+1)

(n+1)^3-n^3=3n^2+3n+1 得到 n^2=[(n+1)^3-n^3]/3-n-1/3
1²+2²+3²+4²+······+n²=[（n+1)^3-1^3]/3-(1+2+3+...+n)-n/3=n(n+1)(2n+1)/6

1²+2²+3²+4²+······+n²=[n(n+1)(2n+1)]/6