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.

Using the identity from Pascal's Triangle, we get
$$
\begin{align}
\sum_{k=0}^n(-1)^k\binom{k+a}{b}
&=\sum_{k=0}^n(-1)^k\left[\binom{k+a+1}{b}-\binom{k+a}{b-1}\right]\\
&=\sum_{k=1}^{n+1}(-1)^{k-1}\left[\binom{k+a}{b}-\binom{k+a-1}{b-1}\right]\\
&=\frac12\left(\binom{a}{b}+(-1)^n\binom{n+a+1}{b}-\sum_{k=1}^{n+1}(-1)^{k-1}\binom{k+a-1}{b-1}\right)\\
&=\frac12\left(\binom{a}{b}+(-1)^n\binom{n+a+1}{b}-\sum_{k=0}^n(-1)^k\binom{k+a}{b-1}\right)\tag1
\end{align}
$$
Multiplying by $(-2)^b$, we get
$$
\begin{align}
(-2)^b\sum_{k=0}^n(-1)^k\binom{k+a}{b}
&=\frac12(-2)^b\left[\binom{a}{b}+(-1)^n\binom{n+a+1}{b}\right]\\
&+(-2)^{b-1}\sum_{k=0}^n(-1)^k\binom{k+a}{b-1}\\
&=\frac12\sum_{j=1}^b(-2)^j\binom{a}{j}+\frac{(-1)^n}2\sum_{j=1}^b(-2)^j\binom{n+a+1}{j}\\
&+[n\equiv0\bmod2]\\
&=\bbox[5px,border:2px solid #C0A000]{\frac12\sum_{j=0}^b(-2)^j\binom{a}{j}+\frac{(-1)^n}2\sum_{j=0}^b(-2)^j\binom{n+a+1}{j}}\tag2
\end{align}
$$
Setting $a=b=2$ yields
$$
\begin{align}
\sum_{k=0}^n(-1)^k\binom{k+2}{2}
&=\frac18\left[\binom{2}{0}-2\binom{2}{1}+4\binom{2}{2}\right]\\
&+\frac{(-1)^n}8\left[\binom{n+3}{0}-2\binom{n+3}{1}+4\binom{n+3}{2}\right]\\[6pt]
&=\frac18+\frac{(-1)^n}8\left(2n^2+8n+7\right)\tag3
\end{align}
$$
Setting $n=20$ yields
$$
\begin{align}
\sum_{k=0}^{20}(-1)^k\binom{k+2}{2}
&=\frac18+\frac{(-1)^{20}}8\left(2\cdot20^2+8\cdot20+7\right)\\[6pt]
&=121\tag4
\end{align}
$$
