# Proving $\sum_{k=0}^n\binom{2n}{2k} = 2^{2n-1}$ [duplicate]

I'm undergraduate student of mathematics. I need to prove: $$\sum_{k=0}^{n} \binom{2n}{2k}= 2^{2n-1}$$ Can you please help me

• Oct 19, 2013 at 11:57

Expand $\dfrac{(1+x)^{2n}+(1-x)^{2n}}2$ and then plug in $x=1$.

• Sum is up to $\large n$. Oct 19, 2013 at 15:47
• @Felix but the lower index in the binomial coefficient is $2k$. Oct 19, 2013 at 15:58
• @RyanReich $0$k. I got it. Thanks. Oct 19, 2013 at 16:08

Hint: Consider $$(1+x)^{2n} = \sum_{j=0}^{2n} {2n \choose j}x^j$$

1. Plug in $x=-1$ : Divide the above expression into those $j$'s that are even, and those that are odd. Those sums must equal each other.

2. Plug in $x=1$ : You get $2^{2n}$ on the LHS; while the RHS is a sum of the even and odd sums from (1).

$$\sum_{k=0}^{n} \binom{2n}{2k} = \sum_{k=0}^n \left[\binom{2n - 1}{2k} + \binom{2n - 1}{2k - 1}\right] = \sum_{k=-1}^{2n} \binom{2n - 1}{k} = \sum_{k=0}^{2n - 1} \binom{2n - 1}{k} = 2^{2n-1}$$

• Quite clear. Up Vote $0$k. Oct 19, 2013 at 15:49