What is the Result of this Factorial Identity?

Question

Provide a solution for the following sum:

(c) $$\sum\limits_{i=0}^n \binom{2n}{2i} $$ Hint: use this identity:

(b) $$ \binom{n}{r} = \binom{n-1}{r-1} + \binom{n-1}{r}$$

Could you help me with this problem?

This is what I have tried:

$$\sum\limits_{i=0}^n \binom{2n}{2i} = \binom{2n-1}{2i-1} + \binom{2n-1}{2i} \\ = \frac{(2n-1)!}{(2i-1)! [(2n-1)-(2i-1)]!} + \frac{(2n-1)!}{(2i)![(2n-1)-2i]!} \\ = \frac{(2n-1)!}{(2i-1)! (2n-2i)!} + \frac{(2n-1)!}{(2i)!(2n-2i-1)!} \\ = \frac{2i(2n-1)! + (2n-1)!(2n-2i)}{(2i)!(2n-2i)!} \\ = \frac{(2n-1)!(2n)}{(2i)(2n-2i)!}$$

Answer 1

For convenience let $\binom{n}{k}$ be defined for $k\in\mathbb{Z}$ with $\binom{n}{k}:=0$ if $k\notin\left\{ 0,1,\dots,n\right\} $.

Then:

$$\sum_{i=0}^{n}\binom{2n}{2i}=\sum_{i\in\mathbb{Z}}\binom{2n}{2i}=\sum_{i\in\mathbb{Z}}\left[\binom{2n-1}{2i-1}+\binom{2n-1}{2i}\right]=\sum_{i\in\mathbb{Z}}\binom{2n-1}{i}=2^{2n-1}$$

Answer 2

Use exercise 56(b) and induction $$\sum_{i=0}^{n}\binom{2n}{2i}=\binom{2n}{0}+\binom{2n}{2}+\binom{2n}{4}+...+\binom{2n}{2n}=2^{2n-1}$$