请问普洛尼克数列，前n项的和是多少？有公式吗？

1*2+2*3+3*4+....+n(n+1)

=..........+(n^2+n)

=(1+2+3+...+n)+(1^2+2^2+3^2+...+n^2)

=(1/2)n(n+1)+(1/6)n(n+1)(2n+1)

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

1*2 + 2*3 + ... + n(n+1)
= 1+1^2 + 2+2^2 + ... + n+n^2
= 1+2+...+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

另一种算法：
令n=2k,则
(1•2)+(2•3)+(3•4)+(4•5)+(5•6)+(6•7)+……+(n-1)n+n(n+ 1)
= (1•2)+(2•3)+(3•4)+(4•5)+(5•6)+(6•7)+……+(2k-1)(2k)+(2k)(2k + 1)
= 2•（1+3）+ 4•（3+5）+ 6•（5+7）+……(2k)•[(2k-1)+(2k+1)]
= 2•4 + 4•（4×2)+ 6•（4×3) +……+2k•4k
=∑8k^2
= 8•(1^2+2^2+……+k^2)
= 8•[k(k+1)(2k+1)/6]
= (4/3)•k(k+1)(2k+1)
= (2k)•(2k+1)•(2k+2)/3
= n(n+1)(n+2)/3
;

当n=2k+1时,即2k=n-1;
(1•2)+(2•3)+(3•4)+(4•5)+(5•6