平方和公式怎么推导？

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)

(a+b)^2=(a+b)(a+b)=a(a+b)+b(a+b)=a^2+ab+ba+b^2=a^2+2ab+b^2 (a+b)^3=(a+b)(a^2+2ab+b^2)=a(a^2+2ab+b^2)+b(a^2+2ab+b^2)=a^3+2a^2b+ab^2+a^2b+2ab^2+b^3=a^3+3a^2b+3ab^2+b^3....