1×2^2+2×3^2+3×4^2+...+n×(n+1)^2=n×(n+1)×(3n^2+11n+10)/12,用数学归纳法证明

1×2^2+2×3^2+3×4^2+...+n×(n+1)^2=n×(n+1)×(3n^2+11n+10)/12；1×2^2+2×3^2+3×4^2+...+n×(n+1)^2+(n+1)（n+2)^2=(n+1)×(n+2)×(3(n+1)^2+11(n+1)+10)/12...OK

当n=1时,左边=4,右边=4,等式成立
假设 n=k时,
1×2^2+2×3^2+3×4^2+...+k×(k+1)^2=k×(k+1)×(3k^2+11k+10)/12
当 n=k+1时,
左边 = 1×2^2+2×3^2+3×4^2+...+k×(k+1)^2 + (k+1)(k+2)^2
= k×(k+1)×(3k^2+11k+10)/12 + (k+1)(k+2)^2
= k×(k+1)×(k+2)(3k+5)/12 + (k+1)(k+2)^2
= (k+1)(k+2) [ k(3k+5) + 12(k+2) ] / 12
= (k+1)(k+2) [ 3k^2 + 17k + 24 ] / 12
= (k+1)(k+2) [ 3(k+1)^2 + 11(k+1) + 10 ] / 12
所以 n=k+1时也成立
综上,原式成立