Calculate $\sum\limits_{k=0}^{20}(-1)^k\binom{k+2}2$ without calculating each term separately Is it possible to calculate $\sum_{k=0}^{20}(-1)^k\binom{k+2}2$ without calculating each term separately?
The original question was find the number of solutions to $2x+y+z=20$ which I calculated to be the coefficient of $x^{20}$ in $(1+x^2+x^4\dots)(1+x+x^2\dots)^2$ which simplified to the term above.
I know $\sum_{k=0}^{20}\binom{k+2}{2}=\binom{23}3$ but the $(-1)^k$ is ruining things.

Alternatively:
$$\begin{align}\sum_{k=0}^{20}(-1)^k\binom{k+2}2=&\sum_{k=0}^{20}\binom{k+2}2-2\cdot \sum_{k=0}^{9}\binom{2k+3}2=\\
&{23\choose 3}-\sum_{k=0}^9 (2k+3)(k+1)=\\
&1771-2\cdot 825=121.\end{align}$$
