# Is this true that $\sum_{v=0}^k (-1)^v{a+v-1 \choose v}{b+k-v-1 \choose k-v}$ is the coefficient of $t^k$ in $(\frac{1}{1+t})^a(\frac{1}{1-t})^b$

I was reading a paper, in which the author assumed that

$$\sum_{v=0}^k (-1)^v{a+v-1 \choose v}{b+k-v-1 \choose k-v}$$ is the coefficient of $$t^k$$ in $$\left(\frac{1}{1+t}\right)^a\left(\frac{1}{1-t}\right)^b$$ when |t|<1

I tried some examples and it seems to be true. But I wonder how can I prove (disprove) it?

Expanding using the binomial theorem and applying upper negation gives\begin{align} A=\left(\frac 1{1+t}\right)^a=\sum_{r=0}^\infty\binom{-a}r t^r =\sum_{r=0}^\infty \underbrace{(-1)^r\binom{a+r-1}{r}t^r}_{f(r)}\\ B=\left(\frac 1{1+t}\right)^b=\sum_{r=0}^\infty\binom{-b}r t^r= \sum_{r=0}^\infty \underbrace{(-1)^r\binom{b+r-1}{r}t^r}_{g(r)}\end{align} Using the Cauchy Product we have \begin{align} A\cdot B=\left(\frac 1{1+t}\right)^a \left(\frac 1{1+t}\right)^b &=\sum_{k=0}^\infty\sum_{v=0}^k \;f(v)\cdot g(k-v)\\ &=\sum_{k=0}^\infty\sum_{v=0}^k (-1)^v\binom{a+v-1}vt^v\cdot (-1)^{k-v}\binom{b+k-v-1}{k-v}t^{k-v}\\ &=\sum_{k=0}^\infty\color{red}{\underbrace{\sum_{v=0}^k (-1)^k\binom{a+v-1}v\binom{b+k-v-1}{k-v}}_\text{coefficient of t^k}}\;\; t^k \end{align}