<s>[INST] Sum of the sum of the sum of the first $n$ natural numbers I have here another problem of mine, which I couldn't manage to solve.

Given that: $$x_n = 1 + 2 + \dots + n \\ y_n = x_1 + x_2 + \dots + x_n
 \\ z_n = y_1 + y_2 + \dots + y_n $$
Find $z_{20}$.

I know the answer but I'm having a hard time reaching it. I recognized that $x_n$ is obviously $\dfrac{n(n + 1)}{2}$, but how to express the other two in a closed form to allow the calculation? I even tried writing the relations in a recursive way, without success. Is there an easy way to solve this one? I was not allowed to use calculators.  
Thanks,
rubik
 [/INST] Here is a slightly longer method than Brian M. Scott's, relying on knowing closed forms for $n$, $n^{2}$ and $n^{3}$.
You have: $$x_{n}=\frac{n(n+1)}{2}=\frac{1}{2}n^{2}+\frac{1}{2}n$$
Therefore: $$\begin{align}y_{n}&=\sum_{i=1}^{n}{\left(\frac{1}{2}i^{2}+\frac{1}{2}i\right)}=\frac{1}{2}\left(\sum_{i=1}^{n}i^{2}+\sum_{i=1}^{n}{i}\right)=\frac{1}{2}\left(\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}\right) \\ &= \frac{1}{2}\left(\frac{1}{3}n^{3}+n^{2}+\frac{2}{3}n\right) \end{align}$$
Therefore:
$$\begin{align}z_{n}&=\sum_{i=1}^{n}{\left(\frac{1}{6}n^{3}+\frac{1}{2}n^{2}+\frac{1}{3}n\right)}=\frac{1}{6}\sum_{i=1}^{n}n^{3}+\frac{1}{2}\sum_{i=1}^{n}n^{2}+\frac{1}{3}\sum_{i=1}^{n}{n} \\ &= \frac{1}{6}\cdot\frac{n^{2}(n+1)^{2}}{4}+\frac{1}{2}\cdot\frac{n(n+1)(2n+1)}{6}+\frac{1}{3}\cdot\frac{n(n+1)}{2} \\ &= \frac{1}{24}n(n+1)(n+2)(n+3)\end{align}$$
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