Question

求详细的证明过程～ 优先采纳高清规整手写的答案

Answer 1

证明：
设s=1²+2²+3²+……+n²
引用公式：(a+b)³=a³+3a²b+3ab²+b³
1³ = 1³
2³ = (1+1)³=1³+3*1²*1+3*1*1²+1³
3³ = (1+2)³=1³+3*1²*2+3*1*2²+2³
……
n³=(1+n-1)³= 1³+3*1²*(n-1)+3*1*(n-1)²+(n-1)³
(1+n)³ = 1³+3*1²*n+3*1*n²+n³
以上等式最左边全部加起来，等于最右边全部加起来：
1³+2³+3³+……+n³+(1+n)³=(1+n)*1 + 3*1²*(1+2+3+……+n) + 3*1*(1²+2²+3²+……+n²)+1³+2³+……+n³
左右消去1³+2³+3³+……+n³，求和1+2+3+……+n
(1+n)³ = (1+n) +3(1+n)*n/2 + 3s
3s = (1+n)³ - (1+n) -3(1+n)*n/2
6s = 2(1+n)³ - 2(1+n) - 3(1+n)*n
6s = (1+n)[2(1+n)²-2-3n)
6s = (1+n)(2n²+n) = (1+n)n(2n+1)
所以: s = n(n+1)(2n+1)/6