prove by inductive step

Question

I have some problem to prove this statement by the Principle of mathematical Induction.

$$\sum_{i=0}^{n} \binom{n}{i} = 2^n.$$ So I begin with the base step. For $n=0$, $$\sum_{i=0}^{0} \binom{0}{i} = 2^0 =1.$$

Now could you help me to show the inductive step?

Thank you in advance!

Answer 1

Hints: under the agreement that $\;\binom nk=0\;$ if $\;k>n\;\;or\;\;k<0\;$ , we get

$$\begin{align}&\bullet\;\;\binom{n+1}{k}=\binom n{k-1}+\binom nk\\{}\\ &\bullet\;\;\sum_{k=0}^{n+1}\binom{n+1}k=\sum_{k=0}^{n+1}\left[\binom n{k-1}+\binom nk\right]=\sum_{k=0}^n\binom nk+\sum_{k=0}^n\binom nk=2^n+2^n=2^{n+1}\end{align}$$