How can i prove that $\displaystyle\sum_{k=0}^{n}{2n\choose 2k}=2^{2n-1}$
I tried using induction and pascal's identity but it didn't help me.
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We know:$$ (1+x)^{2n}=\sum_{j=0}^{2n}{2n \choose j}x^j$$ Now put $x=1$ and and $x=-1$ in turn and add,
$\displaystyle 2^{2n}=(1+1)^{2n}+(1-1)^{2n}=\sum_{j=0}^{2n}{2n \choose j}+\sum_{j=0}^{2n}{2n \choose j}(-1)^j=\sum_{j=0}^{n}{2n \choose 2j}+\sum_{j=0}^{n-1}{2n \choose 2j+1}+\sum_{j=0}^{n}{2n \choose 2j}-\sum_{j=0}^{n-1}{2n \choose 2j+1}=2\sum_{j=0}^{n}{2n \choose 2j}$
$\displaystyle \Rightarrow 2^{2n-1}=\sum_{j=0}^{n}{2n \choose 2j}$
amWhy
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Abhra Abir Kundu
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Can you please explain this transition? $\sum_{j=0}^{2n}{2n \choose j}+\sum_{j=0}^{2n}{2n \choose j}(-1)^j=2\sum_{j=0}^{n}{2n \choose 2j}$ – Dan Barzilay Jul 09 '13 at 16:23
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1@DanBarzilay now I think it will be clear. – Abhra Abir Kundu Jul 09 '13 at 17:35
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The number of ways to select an even subset out of a $2n$-set, is th enumber of ways to select an arbitrary subset among the first $2n-1$ elements and then either take the last element or not to ensure an even number.
Hagen von Eitzen
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