Find the closed form for the double sum $\sum_{i=1}^n \sum_{j=i}^n 2j$

Question

This is what i get:

$$n^3 + n^2 - n(n+1)(2n+1)/6 + n(n+1)/2$$

When I simplify :

I get : $(1/3)n(2n^2+3n+1)$

Is anyone else getting the same result.

Answer 1

Your answer is correct.

\begin{align*} \sum_{i=1}^n\sum_{j=i}^n2j &=\sum_{j=1}^n\sum_{i=1}^j2j\\ &=\sum_{j=1}^n\left(2j\sum_{i=1}^j1\right)\\ &=\sum_{i=1}^n2j^2\\ &=2\cdot\frac{n(n+1)(2n+1)}{6}\\ &=\frac{n(n+1)(2n+1)}{3} \end{align*}

Answer 2

[Alpha](http://www.wolframalpha.com/input/?i=sum(sum(2j,+%7Bj,i,n%7D),%7Bi,1,n%7D) agrees with you.......