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
Solution 1:
Expand $\dfrac{(1+x)^{2n}+(1-x)^{2n}}2$ and then plug in $x=1$.
Solution 2:
Hint: Consider $$ (1+x)^{2n} = \sum_{j=0}^{2n} {2n \choose j}x^j $$
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.
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).