Find $\sum_{k=0}^{n}{{2n+1}\choose{k}}$.

Question

Calculate $\displaystyle\sum_{k=0}^{n}{{2n+1}\choose{k}}$.

What I know:
I know that $\displaystyle\sum_{k=0}^{n}{{n}\choose{k}}=2^n$. I've also figured that what I have to find is all subsets of size $\leq$ n in a set of size $2n+1$. So: $\displaystyle\sum_{k=0}^{n}{{n}\choose{k}}=\displaystyle2^{2n+1}-\sum_{k=n}^{2n+1}{{2n+1}\choose{k}}$. Beyond this, I'm lost.

Thanks for the help!

Answer 1

Hint: $${2n+1 \choose{k}} = {2n+1 \choose{2n+1-k}}$$

Answer 2

As $\binom mr=\binom m{m-r},$

$$2\sum_{k=0}^n\binom{2n+1}k=\sum_{k=0}^{2n+1}\binom{2n+1}k=(1+1)^{2n+1}$$

See also: Alternating sum of binomial coefficients: given $n \in \mathbb N$, prove $\sum^n_{k=0}(-1)^k {n \choose k} = 0$

Answer 3

We know that $\sum\limits_{k = 0}^{2n + 1} \binom{2n+1}{k} = 2^{2n + 1}$.

We also know that $\sum\limits_{k = 0}^n \binom{2n + 1}{k} = \sum\limits_{k = n + 1}^{2n + 1} \binom{2n + 1}{2n + 1 - k}$ (this is by taking the map $k \mapsto 2n + 1 - k$).

And we also have $\binom{n}{k} = \binom{n}{n - k}$ for all $n$, $k$, so in particular, we have $\sum\limits_{k = n + 1}^{2n + 1} \binom{2n + 1}{2n + 1 - k} = \sum\limits_{k = n + 1}^{2n + 1} \binom{2n + 1}{k}$.

This gives us $$\begin{align}2^{2n + 1} &= \sum\limits_{k = 0}^{2n + 1} \binom{2n+1}{k} \\&= \sum\limits_{k = 0}^n \binom{2n + 1}{k} + \sum\limits_{k = n + 1}^{2n + 1} \binom{2n + 1}{k} \\&= 2 \cdot \sum\limits_{k = 0}^n \binom{2n + 1}{k}.\end{align}$$

Therefore, $$\sum\limits_{k = 0}^n \binom{2n + 1}{k} = 2^{2n} = 4^n.$$