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

Question

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

Related : http://math.stackexchange.com/questions/530676/prove-sum-i-0n-binom2n1i-22n — lab bhattacharjee, Oct 19 '13 at 11:57

score 16 · Answer 1 · answered Oct 19 '13 at 10:09

16

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

answered Oct 19 '13 at 10:09

bof

6,156

Sum is up to $\large n$. – Felix Marin Oct 19 '13 at 15:47
@Felix but the lower index in the binomial coefficient is $2k$. – Ryan Reich Oct 19 '13 at 15:58
@RyanReich $0$k. I got it. Thanks. – Felix Marin Oct 19 '13 at 16:08

score 7 · Answer 2 · answered Oct 19 '13 at 10:08

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).

score 4 · Answer 3 · edited Oct 09 '22 at 23:11

4

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

edited Oct 09 '22 at 23:11

B Hos

123

answered Oct 19 '13 at 13:46

RiaD

1,272

Quite clear. Up Vote $0$k. – Felix Marin Oct 19 '13 at 15:49

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

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