$$ \sum_{i=0}^n {2n \choose 2i} = 2^{2n-1} $$
I know what this sum is supposed to equal. I also have a hint that I am supposed to use ${n \choose r} = {n-1 \choose r-1} + {n-1 \choose r}$
I was just wondering if someone could help me with where to start.
Do you start with:
$$ {2n-1 \choose 2i-1} + {2n-1 \choose 2i}$$ $$={(2n-1)! \over (2i-1)!(2n-1-(2i-1)!}+{(2n-1)! \over 2i!(2n-1-2i)!}$$
Hint: $$\sum_{i\text{ is odd}} {2n \choose i} = \sum_{i \text{ is even}}{ 2n\choose i}$$ and $$ \sum_i {2n \choose i} = 2^{2n} $$
We have $$2^n=(1+1)^n=\sum_{k=0}^{n}\binom{n}{k}$$and$$0=(1-1)^n=\sum_{k=0}^{n}\binom{n}{k}(-1)^k.$$ Adding these gives you $$2^n=2\left(\binom{n}{0}+\binom{n}{2}+\binom{n}{4}+\cdots\right),$$ i.e. $$\binom{n}{0}+\binom{n}{2}+\binom{n}{4}+\cdots=2^{n-1}.$$
Now $n\to 2n$.
Recommendation: use the Binomial Theorem on $$ (1+x)^{2n} + (1-x)^{2n}, $$ then set $x=1$.
I believe there is a small typo in the statement: perhaps you meant $$ \sum_{i=0}^n \binom{2n}{2i}=2^{2n-1}.$$
In order to prove this, you need to know that $$ \sum_{j=0}^m (-1)^j\binom{m}{j}=0$$ and $$ \sum_{j=0}^m\binom{m}{j}=2^m.$$ [This holds true because of the binomial theorem applied to $(1-1)^m=0$ and $(1+1)^m=2^m$, respectively. Alternatively you can look for some combinatorial interpretation.]
The first equation gives you $$S=\sum_{j\ \mathrm{even}} \binom{m}{j}= \sum_{j\ \mathrm{odd}} \binom{m}{j}$$
while the second equation gives you $2S=2^m$, i.e. $S=2^{m-1}$. Setting $m=2n$ yields what you want.
The sum $\sum_{i=0}^n{2n\choose 2i}$ counts the sets in ${\cal P}:={\cal P}\bigl([2n]\bigr)$ having an even number of elements. The map $\iota: \>A\mapsto A\triangle\{1\}$ (symmetric difference) is an involution of ${\cal P}$ interchanging the parity of the affected sets. It follows that the sum in question is $={1\over2}|{\cal P}|=2^{2n-1}$.
