Factorise confusion

Question

$$\dfrac{1 - (1 - a)^{n + 1} + a(1 - a)^{n + 1}}{a} = \dfrac{1 - (1 - a)(1 - a)^{n + 1}}{a}$$

Can someone please explain how this simplification works? thank you

Answer 1

Let $u = (1-a)^{n+1}$. Then we have $$\frac{1 - u + au}a = \frac{1 + (-1 + a)u}a = \frac{1 - (1-a)u}a.$$

Answer 2

$\dfrac{1 - \color{red}{(1 - a)^{n + 1}} + a\color{red}{(1 - a)^{n + 1}}}{a} =$

$\dfrac{1 - [\color{red}{(1 - a)^{n + 1}} - a\color{red}{(1 - a)^{n + 1}}]}{a} =$

$\dfrac{1 - [1\cdot\color{red}{(1 - a)^{n + 1}} - a\color{red}{(1 - a)^{n + 1}}]}{a} =$

$ \dfrac{1 - (1 - a)\color{red}{(1 - a)^{n + 1}}}{a}$

And going one step further:

$ \dfrac{1 - \color{blue}{(1 - a)}\color{blue}{(1 - a)}^{n + 1}}{a}= \dfrac{1 - \color{blue}{(1 - a)}^{n + 2}}{a}$