<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] $$2y_n=2\sum_{1\le r\le n}x_1=\sum_{1\le r\le n}(r^2+r)=\frac{n(n+1)(2n+1)}6+\frac{n(n+1)}2=\frac{n(n+1)(n+2)}3$$
$\implies 6y_n=n^3+3n^2+2n$
$$6z_n=6\sum_{1\le r\le n}y_r=\sum_{1\le r\le n}(r^3+3r^2+2r)$$
$$=\left(\frac{n(n+1)}2\right)^2+3\frac{n(n+1)(2n+1)}6+2\frac{n(n+1)}2$$
$$=\frac{n(n+1)\{3n(n+1)+6(2n+1)+12\}}{12}=\frac{n(n+1)\{3n^2+15n+18\}}{12}=\frac{n(n+1)(n+2)(n+3)}4$$
So, $$z_n=\frac{n(n+1)(n+2)(n+3)}{24}$$
So, $$z_{20}=\frac{20(20+1)(20+2)(20+3)}{24}=5\cdot7\cdot11\cdot 23$$
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