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

There’s a very easy way if you know some basic facts about binomial coefficients. You have $x_n=\binom{n+1}2$, so 
$$y_n=\sum_{k=1}^nx_k=\sum_{k=1}^n\binom{k+1}2=\binom{n+2}3$$
and
$$z_n=\sum_{k=1}^ny_n=\sum_{k=1}^n\binom{k+2}3=\binom{n+3}4\;,$$
and $$z_{20}=\binom{23}4=\frac{23\cdot22\cdot21\cdot20}{4\cdot3\cdot2}=23\cdot11\cdot7\cdot5=253\cdot7\cdot5=1771\cdot5=8855$$
(where the arithmetic was done in my head as I was typing).
Without that knowledge (or something comparable, like the formulas for the sums of consecutive squares and cubes) it would appear to be a pretty hard problem. The specific identity that I’m using is 
$$\sum_{k=m}^n\binom{k}m=\binom{n+1}{m+1}\;.$$
