# Simplify summation into closed form

Have no idea where to go. A friend suggested taking the difference of Gauss sums, not sure what that means:

$$\sum_{i = 1}^{n}\sum_{j = i}^{2i}j$$

Recall that $s(n) = \sum_{k=1}^n k = n(n+1)/2$ and note that $$\sum_{j=i}^{2i}j = \sum_{j=1}^{2i} j - \sum_{j=1}^{i-1} j = s(2i) - s(i-1)$$
• @user7252850 you add and subtract the terms from 1 to $i-1$ (i.e. $s(i-1)$) and the result is all terms from 1 to $2i$... – gt6989b Sep 8 '17 at 1:02