Showing the binomial identity of $\sum_{k=0}^{2n}\binom{2n}{k}=\sum_{k=0}^n \binom{2n}{k}+\sum_{k=1}^n \binom{2n}{n+k}$

Question

I want to show the binomial $$\sum_{k=0}^{2n}\binom{2n}{k}=\sum_{k=0}^n \binom{2n}{k}+\sum_{k=1}^n \binom{2n}{n+k}$$

I have tried the following $$\sum_{k=1}^n \binom{2n}{n+k}=\sum_{k=1}^n \binom{2n}{n-k}$$

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

How do I obtain the summation index of $2n$ from the above two summations?

Answer 1

You don't need any identity. It's just simply writing

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

and then noticing that in the second sum the terms can be written as $\binom{2n}{n+j}$ where $j$ runs from $1$ to $n$.