Answer
$0$, see explanations.
Work Step by Step
Step 1. Find the pattern of the sums $S_n$:
$S_1=\begin{pmatrix} 1\\0 \end{pmatrix}-\begin{pmatrix} 1\\1 \end{pmatrix}=0$,
$S_2=\begin{pmatrix} 2\\0 \end{pmatrix}-\begin{pmatrix} 2\\1 \end{pmatrix}+\begin{pmatrix} 2\\2 \end{pmatrix}=1-2+1=0$,
$S_3=\begin{pmatrix} 3\\0 \end{pmatrix}-\begin{pmatrix} 3\\1 \end{pmatrix}+\begin{pmatrix} 3\\2 \end{pmatrix}-\begin{pmatrix} 3\\3 \end{pmatrix}=1-3+3-1=0$
Step 2. We can generalize the findings as:
$S_n=\begin{pmatrix} n\\0 \end{pmatrix}-\begin{pmatrix} n\\1 \end{pmatrix}+\begin{pmatrix} n\\2 \end{pmatrix}- ... + (-1)^n \begin{pmatrix} n\\n \end{pmatrix}=0$
Step 3. We can prove the above finding by expanding $(1-1)^n$ using the Binomial Theorem (omitting all $1^r$ terms):
$(1-1)^n=\begin{pmatrix} n\\0 \end{pmatrix}-\begin{pmatrix} n\\1 \end{pmatrix}+\begin{pmatrix} n\\2 \end{pmatrix}- ... + (-1)^n \begin{pmatrix} n\\n \end{pmatrix}=S_n$.
Thus $S_n=(1-1)^n=0$
