<s>[INST] Nested Summation I have the following nested sum :
$$\sum_{i=1}^{n}\sum_{j=1}^{i}\sum_{k=1}^{j}x = x+1$$
I don't have a clue how to solve this one, can somebody help me?
Thanks in advance.
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 [/INST] I will use the rules found on this page.  
First, note that $x$ is not in the base or limit of any of the sums, so it can be "pulled out":
$$x\sum_{i=1}^n\sum_{j=1}^i\sum_{k=1}^j 1= x+1$$
Work from inside to outside, simplifying the sum:
$$\sum_{k=1}^j 1 = j$$
$$\sum_{j=1}^ij = \frac{i(i+1)}{2}=\frac{1}{2}\cdot\left(i^2 + i\right)$$
$$\sum_{i=1}^n\frac{1}{2}\cdot\left(i^2 + i\right) = \frac{1}{2}\left(\sum_{i=1}^ni^2+\sum_{i=1}^ni\right) = \frac{1}{2}\left(\frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}\right)$$
Now we're done, and can plug back in:
$$\frac{x}{2}\left(\frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}\right)=x+1$$
Now, solving for $x$:
$$x\left(\frac{n(n+1)(2n+1)}{12} + \frac{n(n+1)}{4} - 1\right)=1$$
$$
\begin{align*}
x &= \frac{1}{\left(\frac{n(n+1)(2n+1)}{12} + \frac{n(n+1)}{4} - 1\right)} \\
  &= \frac{6}{n^3 + 3 n^2 + 2 n - 6}
\end{align*}
$$
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