<s>[INST] Find the sum using The question is as follows:
Find the sum:
$1\cdot2 + 2\cdot3  + ... +  (n-1)n$
What I have tried so far:
We can write $(n-1)n$  as $\frac{(n+1)!}{(n-1)!}$ which we can also write as  $2\cdot\dbinom{n+1}{2}$ 
I believe it is possible to use the binomial theorem here, setting $a = b = 1$ in $(a+b)^n$. I am not sure how to proceed however.
 [/INST] $$1\cdot2 + 2\cdot3  + ... +  (n-1)n+\color{red}{n(n+1)-n(n+1)}=$$
$$=1(1+1)+2(2+1)+...+(n-1)n+\color{red}{n(n+1)-n(n+1)}=$$
$$=1^2+2^2+...+n^2+1+2+...+n-n(n+1)=$$
$$=\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}-n(n+1)=$$
$$=\frac{n(n+1)(n+2)}{3}-n(n+1)=\frac{n(n^2-1)}{3}$$
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