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.
Hint: Rewrite $k(k+1)$ as $\frac{1}{3}((k+1)^3-k^3)-\frac{1}{3}$. Observe the mass cancellation (telescoping) when we add up.
Another way: As remarked in the post, we want $$2\left(\binom{2}{2}+\binom{3}{2}+\binom{4}{2}+\cdots+\binom{n}{2}\right).$$ We show that the sum of the binomial coefficients is $\binom{n+1}{3}$. That will show that our sum is $2\cdot\binom{n+1}{3}$, which simplifies to $\frac{(n+1)(n)(n-1)}{3}$.
There are $n+1$ different doughnuts in a row. We want to choose $3$ of them for a healthy breakfast. There are $\binom{n+1}{3}$ ways to do the choosing.
Let us count another way. Maybe the leftmost doughnut we choose is the first doughnut. Then there are $\binom{n}{2}$ ways to choose the other $2$.
Maybe the leftmost doughnut we choose is the second one. There are then $\binom{n-1}{2}$ ways to choose the other $2$. Continue. Maybe the leftmost doughnut we choose is the third from the right end. Then there are $\binom{2}{2}$ ways to choose the other $2$. We have shown that $$\binom{n+1}{3}=\binom{n}{2}+\binom{n-1}{2}+\cdots +\binom{2}{2}.$$
$$1\cdot2 + 2\cdot3 + ... + (n-1)n+\color{red}{n(n+1)-n(n+1)}=$$ $$=1(1+1)+2(2+1)+...+(n-1)n+\color{red}{n(n+1)-n(n+1)}=$$ $$=1^2+2^2+...+n^2+1+2+...+n-n(n+1)=$$ $$=\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}-n(n+1)=$$ $$=\frac{n(n+1)(n+2)}{3}-n(n+1)=\frac{n(n^2-1)}{3}$$
