$\sum_{\alpha=0}^A\binom{A}{\alpha}(zb)^{\alpha} c^{A-\alpha} = (bz+c)^A$

Question

Why the following equality holds:$$\sum_{\alpha=0}^A\binom{A}{\alpha}(zb)^{\alpha} c^{A-\alpha} = (bz+c)^A$$ How to prove it?

Answer 1

If you know Pascal's Triangle, where $$ \binom{n-1}{k-1}+\binom{n-1}{k}=\binom{n}{k}\tag1 $$ we can prove the Binomial Theorem via induction.

First, it holds for $A=0$: $$ \begin{align} (bz+c)^0 &=\sum_{a=0}^0\binom{0}{0}(bz)^0c^0\tag2\\[3pt] &=1\tag3 \end{align} $$ Suppose the Theorem holds for $A-1$, then $$ \begin{align} (bz+c)^A &=(bz+c)(bz+c)^{A-1}\tag4\\[6pt] &=(bz+c)\sum_{a=0}^{A-1}\binom{A-1}{a}(zb)^ac^{A-1-a}\tag5\\ &=\sum_{a=0}^{A-1}\binom{A-1}{a}(zb)^{a+1}c^{A-1-a}+\sum_{a=0}^{A-1}\binom{A-1}{a}(zb)^ac^{A-a}\tag6\\ &=\sum_{a=1}^A\binom{A-1}{a-1}(zb)^ac^{A-a}+\sum_{a=0}^{A-1}\binom{A-1}{a}(zb)^ac^{A-a}\tag7\\ &=\sum_{a=0}^A\binom{A-1}{a-1}(zb)^ac^{A-a}+\sum_{a=0}^A\binom{A-1}{a}(zb)^ac^{A-a}\tag8\\ &=\sum_{a=0}^A\binom{A}{a}(zb)^ac^{A-a}\tag9 \end{align} $$ Explanation:
$(4)$: $x^n=x\,x^{n-1}$
$(5)$: the Theorem holds for $A-1$
$(6)$: distribute $(bz+c)\sum=bz\sum+c\sum$
$(7)$: substitute $a\mapsto a-1$ in the left sum
$(8)$: $\binom{A-1}{-1}=\binom{A-1}{A}=0$ so we include those terms
$(9)$: apply $(1)$

Thus, the Theorem holds for $A$.

Therefore, the Theorem holds for all $A\ge0$.

$\large\square$

Answer 2

This is just the binomial formula, there isn't much to prove here.

Binomial theorem