Prove that $\sum_{j=k}^{2k}(-1)^j\binom{n-j}{i}\binom{k}{j-k} =(-1)^k \binom{n-2k}{i-k}$

Question

I would like to prove that $$ \sum_{j=k}^{2k}(-1)^j\binom{n-j}{i}\binom{k}{j-k} = (-1)^k \binom{n-2k}{i-k} $$ Thank you for your help.

Answer 1

Seeking to evaluate

$$\sum_{j=k}^{2k} (-1)^j {n-j\choose q} {k\choose j-k}$$

we start with

$$(-1)^k \sum_{j=0}^{k} (-1)^j {n-j-k\choose q} {k\choose j} \\ = (-1)^k [z^q] (1+z)^{n-k} \sum_{j=0}^{k} (-1)^j {k\choose j} (1+z)^{-j} \\ = (-1)^k [z^q] (1+z)^{n-k} \left(1-\frac{1}{1+z}\right)^k \\ = (-1)^k [z^q] (1+z)^{n-2k} z^k = (-1)^k {n-2k\choose q-k}$$

as claimed.