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?
 A: 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 $$\binom{N+3}{3}-\binom{3}{3} = \binom{N+3}{3}-1$$
A: $$\sum\limits_{k=1}^N \binom{k+2}{2}=\frac{1}{2}\sum\limits_{k=1}^N(k^2+3k+2)=\frac{1}{2}(\sum\limits_{k=1}^Nk^2+\sum\limits_{k=1}^N3k+\sum\limits_{k=1}^N2)$$ $$=\frac{N(N+1)(2N+1)}{12}+\frac{3N(N+1)}{4}+N$$
A: The set of solutions of the set of equations 
$$x + y + z = k  \qquad 1 \leq k \leq N \tag{1}$$
in the non-negative integers is the set of solutions of the inequality
$$x + y + z \leq N \tag{2}$$
in the non-negative integers with the exception of $(0, 0, 0)$.  The number of solutions of the inequality in the non-negative integers is the number of solutions of the equation
$$w + x + y + z = N \tag{3}$$
where $w = N - x - y - z$.  The number of solutions of equation 3 in the non-negative integers is equal to the number of ways three addition signs can be placed in a row of $N$ ones, which is 
$$\binom{N + 3}{3}$$
since we must select which $3$ of the $N + 3$ symbols (three addition signs and $N$ ones) will be addition signs.  Hence, the number of solutions of the set of equations $x + y + z = k$ if $1 \leq k \leq N$ in the non-negative integers is 
$$\binom{N + 3}{3} - 1$$
whence
$$\sum_{k = 1}^{N} \binom{k + 2}{2} = \binom{N + 3}{3} - 1$$ 
