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}$$=1+7+21+35+35=99$
No apparent pattern here
Thank you for your help
We have ${p \choose k} = {p \choose p-k}$, so $$ \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 $2^{p-1} + {p \choose p/2}/2$.