Prove Combinatorical and Algebrical $\displaystyle\sum_{k=0}^\mathbb{n}$ ${2n+1 \choose k}$ = $2^{2n}$ stuck at this question.
My friend told me to use Newton's Binominal formula but I don't understand how -_-
$\displaystyle\sum_{k=0}^\mathbb{n}$ ${2n+1 \choose k}$ = $2^{2n}$
Also , in the combinatorical way I know that $2^{2n}$ is the amount of subsets that exist in a set with 2n elements.
but I don't understand why its equal to the Left-hand-side?
Thank you very much!
 A: Say we have $2n+1$ people. Choose majority of these (can be everyone) to be red team and the rest will be blue team. The number of people in blue team is $k=0,...,n$. The number of possible team assignment is
$$
\sum_{k=0}^{n}{\binom{2n+1}{k}}
$$
Here is another way, first divide the people into two groups. Then the group with more people will be red team. The number of ways to divide $2n+1$ people into two groups is
$$
2^{2n}
$$
A: W.K.T $${n \choose r} = {n \choose n-r}$$
So, $$ {2n+1 \choose 0} = {2n+1 \choose 2n+1}$$
$${2n+1 \choose 1} = {2n+1 \choose 2n}$$
$${2n+1 \choose 2} = {2n+1 \choose 2n-1}$$
and so on until
$${2n+1 \choose n} = {2n+1 \choose n+1}$$
$${2n+1 \choose n+1} = {2n+1 \choose n}$$
Hence,
$$\sum_{k=0}^n {2n+1 \choose k} = \sum_{k=n+1}^ \left(2n+1\right) {2n+1 \choose k} =  
 1/2\sum_{k=0}^ \left(2n+1\right) {2n+1 \choose k}$$
$$ (1+x)^ {2n+1} = C_0 + C_1 X + C_2 X^2 +.... + C_\left(2n+1\right) x^ {2n+1} $$ from binomial expansion
So,
$$ 2^ \left(2n+1\right) = \sum_{k=0}^ \left(2n+1\right) {2n+1 \choose k}$$
$$\sum_{k=0}^n {2n+1 \choose k} =  1/2\sum_{k=0}^ \left(2n+1\right) {2n+1 \choose k} = 1/2* 2^ \left(2n+1\right) = 2^{2n} $$
