Evaluate $$ \sum_{k=0}^n {{2n + 1}\choose {2k + 1}} $$
I'm really stuck on this one, no idea how to progress. My best guess is to somehow get it into the form of $ n\choose k $ and then take that summation and work with that. Or maybe binomial theorem, but I'm very experienced with that. If you could give a breakdown on how to tackle these problems that'd be great!
Best Answer
Hint:
By applying Binomial Theorem to $ (1-1)^{2n+1}=0 $, we have:
$$ \binom{2n+1}{0}-\binom{2n+1}{1}+\binom{2n+1}{2}-\cdots+\binom{2n+1}{2n}-\binom{2n+1}{2n+1}=0 .$$
Move the negative terms to the right-hand side, we have:
$$ \binom{2n+1}{1}+\binom{2n+1}{3}+\cdots+\binom{2n+1}{2n+1}=\binom{2n+1}{0}+\binom{2n+1}{2}+\cdots+\binom{2n+1}{2n} .\tag{A}$$ And applying Binomial Theorem to $ (1+1)^{2n+1}=2^{2n+1} $, we have: $$ \binom{2n+1}{0}+\binom{2n+1}{1}+\binom{2n+1}{2}+\cdots+\binom{2n+1}{2n}+\binom{2n+1}{2n+1}=2^{2n+1} .\tag{B}$$
Further hint:
$$ 2\times (A)=(B) .$$