Geometric Summations that reference two variables

Hello math stackexchange! I'm new here so please correct any formatting mistakes I make / I'm happy to provide more info if needed.

I have a summation of the form $\sum_{i=1}^{n} \sum_{j=1}^{i} (i+j)$, and I'm not too sure how to go about solving this (and by solve, I mean rewrite it into a formula) . So far I've been able to refactor the inner set of brackets into $\sum_{i=1}^{n} \big( \sum_{j=1}^{i} i + \sum_{j=1}^{i} j \big) $. From here though I'm not too sure what to do as my inner summations depend on two variables (the i and j cannot be factored out)

From here I've tried the following:

• Rewrite the inner summations to iterate up to n instead of i. However I'm not too sure if this is a valid way of continuing the problem:

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

• Given that the variable i is in the scope of the inner summations, i can be factored out and then solve the inner summation for j.

$\sum_{i=1}^{n} \sum_{j=1}^{i} (i+j) = \sum_{i=1}^{n} \big( \sum_{j=1}^{i} i + \sum_{j=1}^{i} j \big)$

$= \sum_{i=1}^{n} i \big(\sum_{j=1}^{i} 1 + \sum_{j=1}^{i} j \big) $

This only gets me up to this point, because I don't know a formula to rewrite a nested summation that iterates up to i.

• That lead me to combining both those ideas and ending up with this:

$\sum_{i=1}^{n} \sum_{j=1}^{i} (i+j) = \sum_{i=1}^{n} \sum_{j=i}^{n} (i+j)$ [1]

$= \sum_{i=1}^{n} i \big( \sum_{j=i}^{n} (1) + \sum_{j=i}^{n} (j) \big)$ [2]

$= \sum_{i=1}^{n} i \big(n + \frac{n(n+1)}{2} \big)$ [3]

$=\sum_{i=1}^{n} i n + \sum_{i=1}^{n} i \frac{n(n+1)}{2} $ [4]

$=\frac{1}{2} n^2 (n+1) + \frac{1}{2} \sum_{i=1}^{n} i n(n+1)$ [5]

$=\frac{1}{2} n^2 (n+1) + \frac{1}{2} n^2 (n+1)^2$ [6]

Which simplifies to $\frac{1}{2} n^2 (n+2)(n+1)$. Wolfram's solution was $\frac{1}{2} n(n+1)^2$. I'm wondering which of my assumptions are invalid, or is it an issue to do with my process? Maybe I'm overcomplicating the fact that there is two variables being referenced inside the summations, and this problem could be simplified in a different way?

Cheers

Answer

Correction:

$$\sum_{i=1}^{n} \sum_{j=1}^{i} (i+j) = \sum_{i=1}^{n} \left[i\left( \sum_{j=1}^{i} 1\right) + \sum_{j=1}^{i} j \right]$$

To obtain the answer:
\begin{align}
\sum_{i=1}^{n} \sum_{j=1}^{i} (i+j) &= \sum_{i=1}^{n} \left[i^2 + \frac{i(i+1)}{2} \right] \\
&=\frac32 \sum_{i=1}^n i^2+\frac12 \sum_{i=1}^n i\\
&=\frac32 \frac{n(n+1)(2n+1)}{6} + \frac12\frac{n(n+1)}{2} \\
&= \frac{n(n+1)(2n+2)}4 \\
&= \frac{n(n+1)^2}{2}
\end{align}