Comparing Two Sums with Binomial Coefficients How do I use pascals identity:
$${2n\choose 2k}={2n-1\choose 2k}+{2n-1\choose 2k-1}$$ to prove that
$$\displaystyle\sum_{k=0}^{n}{2n\choose 2k}=\displaystyle\sum_{k=0}^{2n-1}{2n-1\choose k}$$
for every positive integer $n$ ?

Other than Pascal's identity, we just notice that the sums on the right are the same because they cover the same binomial coefficients (red=even, green=odd, and blue=both).
$$
\begin{align}
\sum_{k=0}^n\binom{2n}{2k}
&=\sum_{k=0}^n\color{#C00000}{\binom{2n-1}{2k}}+\color{#00A000}{\binom{2n-1}{2k-1}}\\
&=\sum_{k=0}^{2n-1}\color{#0000FF}{\binom{2n-1}{k}}
\end{align}
$$
Note that $\binom{2n-1}{-1}=\binom{2n-1}{2n}=0$.
