Question

1×2^2+2×3^2+3×4^2+...+n×(n+1)^2=n×(n+1)×(3n^2+11n+10)/12,用数学

Answer 1

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