证明立方和=和的平方

用数学归纳法
①1+2+3+…+n = n(n+1)/2
.
②验证n=1时，1³ = [1(1+1)/2]² = 1
.
③令n=k时，1³+2³+3³+…+k³ = k²(k+1)²/4
试证n = k+1时，
1³+2³+3³+…+k³+(k+1)³
= k²(k+1)²/4+(k+1)³
= (k+1)²[k²/4+(k+1)]
= [(k+1)²/4](k+2)²
= (k+1)²[(k+1)+1]²/4
.
成立。

k³
= k³ - k + k
= (k+1)k(k-1) + k
= ((k+2)(k+1)k(k-1) - (k+1)k(k-1)(k-2))/4 + k
则
1³+2³+……+n³
= ((n+2)(n+1)n(n-1) - 0) / 4 + (1+2+3+……+n)
= (n+2)(n+1)n(n-1)/4 + n(n+1)/2
= n(n+1)/4b · ((n+2)(n-1)+2)
= (n(n+1)/2)².