# Sum of a combination

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

• Your calculation for $p=7$ is wrong. – Robert Israel Jan 13 at 21:14

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 $$2^{p-1} + {p \choose p/2}/2$$.