Sum of $k(k-1)$

Question

Probably very simple question, but I got stuck anyway. Find $$\sum_{k=1}^{n} k(k-1)$$ Attempt

$\sum_{k=1}^{n} k(k-1) = 2\sum_{k=1}^{n-1} \frac{k(k+1)}{2}=2(1+(1+2)+(1+2+3)+...(1+2+...+n-1))$

Answer 1

You can try

$$ \sum_{k=1}^n k^2 - \sum_{k=1}^n k $$

Both are well known results

$$ \color{blue}{\sum_{k=1}^n k^2} - \color{red}{\sum_{k=1}^n k} = \color{blue}{\frac{1}{6} n (n+1) (2 n+1)} - \color{red}{\frac{1}{2} n (n+1)} = \frac{1}{3} n \left(n^2-1\right) $$

Answer 2

Use that: $$k(k-1)=k^2-k$$ Hence $$\sum_{k=1}^{n}{k(k-1)}=\sum_{k=1}^{n}{k^2}-\sum_{k=1}^{n}{k}$$ Then note that: $$\sum_{k=1}^{n}{k^2}=\frac n6(n+1)(2n+1)$$ $$\sum_{k=1}^{n}{k}=\frac n2(n+1)$$ And go from there

Answer 3

Hint. One has $$ k(k-1)=\frac{ k (k-1) (k+1)}{3}-\frac{(k-1) (k-2)k }{3} $$ giving a telescoping sum.

Answer 4

Continuing your line of thought: note it is a sum of triangular numbers: $$\sum_{k=1}^{n} k(k-1) = 2\sum_{k=1}^{n-1} \frac{k(k+1)}{2}=2\sum_{k=1}^{n-1}\frac{(k+1)!}{2!(k-1)!}=2\sum_{k=1}^{n-1} {k+1\choose 2}=\\ 2{(n-1+1)+1\choose 2+1}=2{n+1\choose 3}=2\cdot \frac{(n+1)!}{3!(n-2)!}=\frac{(n-1)n(n+1)}{3}.$$