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.
${}{}$
 A: After a little rearrangement we have
$$\begin{align*}
1+\frac1x&=\sum_{i=1}^n\sum_{j=1}^i\sum_{k=1}^j1\\
&=\sum_{i=1}^n\sum_{j=1}^ij\\
&=\sum_{1=1}^n\frac{i(i+1)}2\\
&=\frac12\left(\sum_{i=1}^ni^2+\sum_{i=1}^ni\right)\\
&=\frac12\left(\frac{n(n+1)(2n+1)}6+\frac{n(n+1)}2\right)\\
&=\frac1{12}\Big(n(n+1)(2n+1)+3n(n+1)\Big)\\
&=\frac{n(n+1)}{12}(2n+4)\\
&=\frac16n(n+1)(n+2)\\
&=\binom{n+2}3\;,
\end{align*}$$
so 
$$\begin{align*}
x&=\frac1{\binom{n+2}3-1}\\
&=\frac6{n(n+1)(n+2)-6}\\
&=\frac6{n^3+3n^2+2n-6}\\
&=\frac6{(n-1)(n^2+4n+6)}\;.
\end{align*}$$
A: 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 $x$:
$$x\left(\frac{n(n+1)(2n+1)}{12} + \frac{n(n+1)}{4} - 1\right)=1$$
$$
\begin{align*}
x &= \frac{1}{\left(\frac{n(n+1)(2n+1)}{12} + \frac{n(n+1)}{4} - 1\right)} \\
  &= \frac{6}{n^3 + 3 n^2 + 2 n - 6}
\end{align*}
$$
A: This question can be solved in a different way.
Your equation is equivalent to $1+\frac1x=\sum_{(i,j,k)\in A}1=|A|$
Where $A={(i,j,k)| 1\leq k  \leq j\leq i\leq n }$
So,$|A|= ^nC_3 +2^nC_2+^nC_1$
This is because, we can choose 3 distinct i,j,k or distinct i with the same j and k, or distinct k with the same i and j, or i=j=k.
So we get $1+\frac1x=^nC_3 +2^nC_2+^nC_1=^{n+2}C_3$
