Prove combinatorically - $\sum_{k=0}^{n} 2^{k}\binom{n}{k} = 3^{n}$ $\sum_{k=0}^{n} 2^{k}\binom{n}{k} = 3^{n}$
I have no idea which story to form, I thought about $n$ students assuming n=4
picking 4 students for a committee with 3 roles. since each students has 3 options.
then on the other side.. $2^{0}\binom{4}{0} + 2^{1}\binom{4}{1} +2^{1}\binom{4}{2} + 2^{3}\binom{4}{3}+2^{4}\binom{4}{4}$ = $2^{0}\binom{4}{4} + 2^{1}\binom{4}{3} +2^{1}\binom{4}{2} + 2^{3}\binom{4}{1}+2^{4}\binom{4}{0}$
but im just stuck here...

The "binomial theorem" says that $(x+ y)^n= \sum_{i=0}^n \begin{pmatrix}n\\ i\end{pmatrix} x^i y^{n-i}$.  If x= 2 and y= 1 that becomes $(2+ 1)^n= \sum_{i=0}^n\begin{pmatrix}n \\i \end{pmatrix} (2^i)(1^{n-i})$ or $3^n= \sum_{i=0}^n \begin{pmatrix}n \\ i\end{pmatrix} 2^i$.
