<s>[INST] sum of solutions of $\{(x,y,z)\mid x+y+z=k\}$, $k = 1,\ldots,N$ What is the sum of non-negative integer solutions of $\{(x,y,z)\mid x+y+z=k\}$, $k = 1,\ldots,N$? 
I know that $\{(x,y,z)\mid x+y+z=k\}$ has $\binom{k+3-1}{3-1}=\binom{k+2}{2}$ non-negative integer solutions. Thus, the answer to the question above is $$\sum_{k=1}^N \binom{k+2}{2}.$$
Can we further simplify the summation?
 [/INST] Here's another elegant way to simplify the expression. See that $\binom{n}{r}$ is the coefficient of $x^r$ in $(1+x)^n$. Hence, your summation can be interpreted as the $$\sum_{k=1}^N \text{coefficient of } x^2 \text{ in } (1+x)^{k+2} = \text{coefficient of } x^2 \text{ in }\sum_{k=1}^N  (1+x)^{k+2}$$ where the last series is a Geometric Progression, which can be summed easily.
$$\sum_{k=1}^N  (1+x)^{k+2} = \sum_{k=3}^{N+2} (1+x)^k = \frac{(1+x)^3((1+x)^N-1)}{1+x-1}=\frac{(1+x)^{N+3}-(1+x)^3}{x}$$
The task is now reduced to finding the coefficient of $x^3$ in $(1+x)^{N+3}-(1+x)^3$ which is $$\binom{N+3}{3}-\binom{3}{3} = \binom{N+3}{3}-1$$
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