How to prove $[x^k] \cdot (1+x)^{a+b} = \sum_{j=0}^k{a\choose j}\cdot{b\choose {k-j}}$

Question

$$[x^k] \cdot (1+x)^{a+b} = \sum_{j=0}^k{a\choose j}\cdot{b\choose {k-j}}$$ where $a,b \in \Bbb R$ and $k \in \Bbb N$. Since, $$(1+x)^{a+b} = \sum_{j=0}^{a+b} {a+b\choose j} \cdot x^j$$

so, $\displaystyle[x^k] \cdot \sum_{j=0}^{a+b} {a+b\choose j} \cdot x^j= {{a+b}\choose k}$

How should I proceed from here? Or am I completely wrong?

Answer 1

Write $$(1+x)^{a+b}=(1+x)^a(1+x)^b=\left(\sum_{j=0}^a\binom ajx^j\right)\left(\sum_{j=0}^b\binom bjx^i\right)$$ and extract the coefficient of $x^k$.