induction with a binomial

Prove $\displaystyle \forall n\geq 0$

$\displaystyle \binom{n}{0}+\binom{n}{1}+\cdots +\binom{n}{n}=2^n$

$\displaystyle p(0): \ \binom{n}{0}=1=2^0=1$

Assume p(k) is true

$\displaystyle p(k): \ \binom{k}{0}+\binom{k}{1}+\cdots +\binom{k}{k}=2^k$

Prove

$\displaystyle p(k+1): \ \binom{k}{0}+\binom{k}{1}+\cdots +\binom{k+1}{k+1}=2^{k+1}$

Simply adding the next binomial term to both sides doesn't help. Any ideas?

Re: induction with a binomial

You're a few "k+1"s short - actually, k of them.

$\displaystyle 2^{k+1} - 2^{k} = 2^{k}\cdot(2-1) = 2^{k}$

Focus on the 3rd term (2 in the bottom of the choose) and see if it makes any sense.

Re: induction with a binomial

Hi dw, it might help if,

$\displaystyle p(k+1): \ \binom{k+1}{0}+\binom{k+1}{1}+\cdots +\binom{k+1}{k+1}=2^{k+1}$