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i cant find an expression for:

$\sum_{k=0}^{\lfloor \frac{p}{2}\rfloor}$ ${p}\choose{k}$

The hint given by the exercise is to do it by seperating the cases where p is odd and even.

What I did so far: When $p$ is even, $\lfloor p/2 \rfloor = p/2$, so all we need is find an expression $\sum_{k=0}^{ \frac{p}{2}}$ ${p}\choose{k}$, which I was not able to do.

When $p$ is odd, there is $q$ so that $p=2q+1$ and $p/2= q +1/2$ which means $\lfloor p/2 \rfloor = q$, and I really dont know what to do with this.

The previous question was to show that functions that are of the form $ax+b$ are both convex and concave.

I thought that I should try to put some values in to look for a pattern and here is what I have : for $p$ even :

$p=2, \sum_{k=0}^{1}$ ${2}\choose{k}$$=1+2=3$

$p=4, \sum_{k=0}^{2}$ ${4}\choose{k}$$=1+6+4=11$

$p=6, \sum_{k=0}^{3}$ ${6}\choose{k}$$=1+6+15+20=42$

no apparent pattern.

when p is odd,

$p=1, \sum_{k=0}^{0}$ ${1}\choose{k}$$=1$

$p=3, \sum_{k=0}^{1}$ ${3}\choose{k}$$=1+3=4$

$p=5, \sum_{k=0}^{2}$ ${5}\choose{k}$$=1+5+10=16$

$p=7, \sum_{k=0}^{4}$ ${7}\choose{k}$$=1+7+21+35+35=99$

No apparent pattern here

Thank you for your help

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    $\begingroup$ Your calculation for $p=7$ is wrong. $\endgroup$ – Robert Israel Jan 13 at 21:14
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We have ${p \choose k} = {p \choose p-k}$, so $$ \sum_{k=0}^{\lfloor p/2 \rfloor} {p \choose k} = \sum_{k = p - \lfloor p/2 \rfloor}^p {p \choose k}$$ Thus if $p$ is odd, $\{0, \ldots, \lfloor p/2 \rfloor \}$ and $\{p - \lfloor p/2 \rfloor, \ldots, p\}$ are disjoint and their union is $\{0, \ldots, p\}$, so using the binomial theorem your sum is $2^{p-1}$. If $p$ is even, $p/2$ is in both sets, so your sum is $2^{p-1} + {p \choose p/2}/2$.

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