Sum of binomial numbers alternating the sign.

Question

Prove $\sum_{p=0}^k {n \choose p}(-1)^p=(-1)^k{n-1 \choose k}$

I tried use the expansion $(1-1)^n=0$ , but I can't find a simetry or binominal relactions.

Answer 1

Because $(-1)^k$ is independent of $p$, you have $$\sum_{p=0}^k {n\choose p}(-1)^k = (-1)^k\cdot\sum_{p=0}^k {n\choose p}$$

Answer 2

Using Pascal rule to prove this -

We know ${{n} \choose {k}}={{n-1} \choose {k-1}}+{{n-1} \choose k}$

So, ${{n-1} \choose k} = {{n} \choose {k}} - {{n-1} \choose {k-1}}$ ...(i)

Using (i) repeatedly

${{n-1} \choose k} = {{n} \choose {k}} - [{n \choose {k-1}} - {{n-1} \choose {k-2}}]$

${{n-1} \choose k} = {{n} \choose {k}} - [{n \choose {k-1}} - [{n \choose {k-2}} - {{n-1} \choose {k-3}}]$

${{n-1} \choose k} = \sum_{p=0}^k {n \choose p}(-1)^{k-p} = (-1)^k \sum_{p=0}^k {n \choose p}(-1)^{-p}$

$ (-1)^k {{n-1} \choose k} = (-1)^{2k} \sum_{p=0}^k {n \choose p}(-1)^{p} \,$ as [$(-1)^{-p} = (-1)^{p}$]

$(-1)^k {{n-1} \choose k} = \sum_{p=0}^k {n \choose p}(-1)^{p}$