Proof for Binomial Expression

Question

I am trying to prove the following equation using the Binomial Theorem:

$\sum_{k=0}^{n} {{2n}\choose{2k}} = \sum_{k=1}^{n} {{2n}\choose{2k-1}}$

I had tried to replace $\binom {2n}{2k}$ with the Pascal Identity: $\binom {2n-1}{2k}+\binom {2n-1}{2k-1}$ but I can't understand how can I simplify it to look like the expression of the right hand side in the equation.

Please help.

Answer 1

From the Binomial Theorem, we have: $$(1+x)^{2n}=\sum_{k=0}^{2n}\binom {2n}{k}x^{k}$$ Now put $x=-1$ in the above expression, you'll get the required result.

Answer 2

You need to replace the RHS with the identity:

$$\sum_{k=1}^{n} \binom{2n}{2k-1}=\sum_{k=1}^{n} \binom{2n-1}{2k-2}+\binom{2n-1}{2k-1}=\sum_{k=0}^{2n-1} \binom{2n-1}{k}=2^{2n-1}$$

Therefore, because $\sum_\limits{k=0}^{2n} \binom{2n}{k}=2^{2n}$, $\sum_\limits{k=0}^{n} \binom{2n}{2k}=2^{2n-1}$ as well.