Sum with binomial coefficient using identity

Question

I want to prove:

$\displaystyle \sum_{k=0}^n (-1)^k \binom{x}{k} = (-1)^n \binom{x-1}{n}$

using: $(1-z)^x \cdot \frac{1}{1-z} = (1-z)^{x-1}$

I know how to do it with induction but i somehow can't figure out how to do it using this identity.

Answer 1

By the binomial theorem, the RHS is the coefficient of $z^n$ in $(1-z)^{x-1}$. For the LHS, use the Cauchy product to compute the same coefficient $c_n$ in the product of $(1-z)^x$ and $1/(1-z)$. Explicitly, the coefficient of $z^n$ in $(1-z)^x$ is $a_n=(-1)^n\binom{x}{n}$, and the coefficient of $z^n$ in $1/(1-z)$ is $b_n=1$, yielding $$c_n = \sum_{k=0}^n a_k b_{n-k} = \sum_{k=0}^n (-1)^k\binom{x}{k} 1 = \sum_{k=0}^n (-1)^k\binom{x}{k}.$$