My textbooks wants me to prove this identity in order to proceed into further questions related to this identity but I tried several hours to prove this but i failed can somebody assist me?

$$\sum_{i=0}^{n} {2n+1 \choose 2i} = 2^{2n}$$

  • $\begingroup$ Can you fix your summation notation, please? There are no limits. $\endgroup$ Commented May 12, 2020 at 1:12
  • 1
    $\begingroup$ How many even-sized subsets are there in a set of odd cardinality? $\endgroup$ Commented May 12, 2020 at 1:30
  • $\begingroup$ It is indeed hard to prove what is not true. Try: $$\sum_{i=0}^n\binom{2n+1}{2i}=4^n$$ $\endgroup$ Commented May 12, 2020 at 1:36

1 Answer 1


Okay. Here are two VERY useful hints!

$\displaystyle \sum_{i=0}^n\binom{n}{i}=2^n$.

Also, $\displaystyle \binom{n}{i} = \binom{n}{n-i}$.

Also, I believe the right hand side should be $2^{2n}$.

  • $\begingroup$ Indeed $\sum_{k=0}^n \binom{2n+1}{2k}=4^n$. $\endgroup$ Commented May 12, 2020 at 1:35

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