Question

1×2+2×3+3×4+......+n(n+1)

给我具体的步骤和说明吧

Answer 1

1×2+2×3+3×4+......+n(n+1)
=(1×1+1)+(2×2+2)+(3×3+3)+......(n×n+n)
=(1^2+2^2+3^2+......n^2)+(1+2+3+......n)
=n*(n+1)*(2*n+1)/6+n(n+1)/2
=n(n+1)(n+2)/3

Answer 2

n(n+1)(n+2)/3

n*(n+1)=n*n+n
所以：1×2+2×3+3×4+4×5+┉┉
=（1*1+2*2+3*3+……）+（1+2+3+……）
前面用前n项的平方和公式，后面有前n项求和公式就可以了。
进一步推导就可得出你要的结论

Answer 3

1^2+2^2+...+n^2=n*(n+1)*(2*n+1)/6
1+2+3+...+n=n(n+1)/2
原式=n(n+1)[(2*n+1)/6+1/2]=n(n+1)(2*n+4)/6
=n(n+1)(n+2)/3

Answer 4

接一楼:
=(1+n2)n/2+(1+n)n/2
=n3/2+n/2+n/2+n2/2
=(n3+n2)/2+n

Answer 5

1*2+2*3+3*4+···+n*(n+1)
=1+2+3+...+n+1^2+2^2+...+n^2
=n(n+1)/2+n(n+1)(2n+1)/6
=n(n+1)(3+2n+1)/6
=n(n+1)(n+2)/3