$\displaystyle \sum_{i=1}^{n-1} i(i+1) = (n(n-1)(n+1))/3.

$ For all integers $\displaystyle n \geq 2

$

Base case: lets n = 2

$\displaystyle \sum_{i=1}^{n-1} i(i+1) = (2(2-1)(2+1))/3.

$

= 2(1)(3)

=2

Inductive step:

$\displaystyle \sum_{i=1}^{(n+1)-1} i(i+1) = $$\displaystyle \sum_{i=1}^{n} i(i+1) + [(n+1)((n+1)+1)]

$

$\displaystyle (n(n-1)(n+1))/3 + [(n+1)(n+2)]

$

$\displaystyle (n(n-1)(n+1)+3(n+1)(n+2))/3

$

$\displaystyle ((n+1)(n^2+2n+6))/3

$

And this is where i am stuck cause I dont really know what to do with the

$\displaystyle n^2+2n+6.

$