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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

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3 Answers 3

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

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

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

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

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