I would like to evaluate $ \sum_{k=0}^{\frac{n}{2}} {n\choose2k}$ - but I can't seem to find a simple expression for this sum. Would appreciate any help.
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1$$(1-1)^n+(1+1)^n=?$$ – lab bhattacharjee Nov 23 '18 at 19:27
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1Is $n{{}}$ even? – Angina Seng Nov 23 '18 at 19:27
3 Answers
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This is the number of subsets of $\{1, \ldots, n\}$ that have even size. If you need a further hint, this question may help.
angryavian
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Hint
$\binom{n}{r}$ is the number of subsets of $[n]=\{1,2,3, \ldots n\}$ of size $r$. So $\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor}\binom{n}{2k}$ is the total number of subsets of $[n]$ with even cardinalities and that would be (about :-)) half the total number of subsets of $[n]$.
Anurag A
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Note that small $n$ needs special treatment (hence the smiley) – Hagen von Eitzen Nov 23 '18 at 20:07
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You should probably write that as $$ \sum_{k=0}^{[n/2]} \binom{n}{2k}$$ otherwise the answer will depend on whether $n$ even of odd. You can equivalently write this as $\sum_{k \text{ even}, n \geq k \geq 0 } \binom{n}{k}$.
The right way to approach this is to consider $(1-1)^n + (1+1)^n$ as suggested in the comment. Expand this using the binomial theorem.
msm
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