Expanding an algebraic expression that is a product of linear expressions to powers

Question

Let $a, b, c ,d, m, n \in \mathbb{Z}$ with $n \geq m$. We have $$(a x+b)^{n-m} = \sum_{i = 0}^{m-n} \binom{n-m}{i} a^{n-m-i} b^i x^{n-m-i} $$ and $$(c x + d)^{m} = \sum_{j = 0}^{m} \binom{m}{j} c^{m-j} d^j x^{m-j} .$$ What is the coefficient of $x^k$ ($k \in \mathbb{Z}, \ 0 \leq k \leq n$) in the expansion of $(a x+b)^{n-m} (c x + d)^{m}$ ?

Answer 1

Setting $$ s_{n-m-i} = \binom{n-m}{i} a^{n-m-i} b^i \quad\text{and}\quad t_{m-j} = \binom{m}{j} c^{m-j} d^j. $$ one gets $$ (ax+b)^{n-m} = \sum_{i = 0}^{m-n} s_{n-m-i}x^{n-m-i}\quad\text{and}\quad (cx + d)^{m} = \sum_{j = 0}^{m} t_{m-j}x^{m-j}. $$ Therefore the coefficient of $x^k$ in the expansion of $(a x+b)^{n-m} (c x + d)^{m}$ is $$ \sum_{(n-m-i) + (m-j) = k\atop 0 \leqslant i \leqslant m-n,\ 0 \leqslant j \leqslant m} s_{n-m-i}t_{m-j} = \sum_{i+j+k = n\atop 0 \leqslant i \leqslant m-n,\ 0 \leqslant j \leqslant m} s_{n-m-i}t_{m-j} $$