Solving Double Summation

Question

I came across the following summation but I don't know how to solve it: $\sum_{i=1}^n\sum_{j=i}^n (j + 1)$ I know how to do double summations, but I never saw double summations that are linked like this before though.

Answer 1

One way I haven't seen in the answers is to reverse the order of the summation

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

which we can use the usual formulas on

$$\frac{n(n+1)(2n+1)}{6}+ \frac{n(n+1)}{2}$$

This ends up being easier because the summand does not depend on $i$ at all, so it's the equivalent of "integrating" a constant.

Answer 2

$\Sigma_{i=1}^n\Sigma_{j=i}^n(j+1)=(\Sigma_{j=1}^n(j+1)+\Sigma_{j=2}^n(j+1)+\cdots\Sigma_{j=n}^n(j+1))$

Answer 3

Hint: rewrite the double summation. $$\sum_{i=1}^n\sum_{j=i}^n(j+1)=\sum_{i=1}^n\left(\sum_{j=1}^n(j+1)-\sum_{j=1}^{i-1}(j+1)\right)$$

Answer 4

For any fixed $i$ we have $$\sum_{j=i}^{n}(j+1)=\frac{(n-i+1)(n+i+2)}{2}.$$ Thus, the sum becomes $$\sum_{i=1}^{n}\frac{(n-i+1)(n-i+2)}{2}=\frac{1}{2}\sum_{i=1}^{n}(n^2+3n-i^2-i+2).$$ Evaluating the sum term by term then gives $$\frac{1}{2}(n^3+3n^2-\frac{n(n+1)(2n+1)}{6}-\frac{n(n+1)}{2}+2n).$$ Simplifying, we get $$\frac{n(n+1)(n+2)}{3}.$$ Feel free to ask if you aren't sure about any of the above steps.

Answer 5

$$\sum_{i=1}^n\sum_{j=i}^n(j+1)=\sum_{i=1}^n\left(\sum_{j=1}^n(j+1)-\sum_{j=1}^{i-1}(j+1)\right)$$ $$\sum_{i=1}^n\sum_{j=i}^n (j + 1) = \sum_{i=1}^n \frac{n.(n+1)}{2} + n - \frac{i(i-1)}{2} -i+1 = \frac{n.(n+1).(n+2)}{3}$$ Now for any general $\sum_{i=1}^n\sum_{j=1}^n f(i,j)$,

• first sum over j taking i as constant in $f(i,j)$ without paying any heed to the summation over i
• once you are done with summation over j, next sum over i as you do normally for a single summation P.S. - It's easier to sum when summation starts with 1. So always try to convert in that form.