Prove $(k+1)\binom{n}{k+1}+k\binom{n}{k}=n\binom{n}{k}$ for integers $0\le k\le n$

Question

Prove $(k+1)\binom{n}{k+1}+k\binom{n}{k}=n\binom{n}{k}$ for integers $0\le k\le n$

I need help, I've been trying to factor all day and can't figure it out.

Answer 1

Before I start my answer, I believe I should state that you should type that instead of using an image.

Proof:

$(k+1)*\binom{n}{k+1} +k * \binom{n}{k}= \frac{k+1}{(k+1)!*(n-k-1)!}+\frac{k*n!}{k!*(n-k)!}$

$= \frac{n!}{k!*(n-k-1)!}+\frac{k*n!}{k!*(n-k)!}=\frac{(n-k)n!}{k!*(n-k)!}+\frac{k*n!}{k!*(n-k)!}$

$=\frac{n(n)!}{k!(n-k)!}$

$=n*\binom{n}{k}$

as required.

Answer 2

$${n \choose k+1} = \frac{n!}{(k+1)!(n-k-1)!} =\frac{n!}{\frac{k+1}{n-k}k!(n-k)!} = \frac{n-k}{k+1}{n \choose k}$$