Simplify $(p)(\frac{1}{2}(2p-1)^n) + (1-p)(-\frac{1}{2}(2p-1)^n)$ to $\frac{1}{2}(2p-1)^{n+1}$ The expression was simplified in the answer to this question. I'm trying to simplify it but I got stuck. Multiplying all the factors and regrouping didn't work, but maybe I'm doing the wrong regrouping.
Also, I don't understand why the $(2p-1)^n$ dropped the exponent in the second line of the same solution.
 A: $$p\cdot (1/2)(2p-1)^n +(1-p) \cdot (-1/2)(2p-1)^n = \\
p \cdot (1/2)(2p-1)^n +(p-1) \cdot(1/2)(2p-1)^{\color{red}{n}}= \\ \left[\frac{p}{2}+\frac{p}{2}-\frac{1}{2}\right](2p-1)^n =\\ (2p-1)\cdot(1/2) (2p-1)^n$$
You are correct that an $n$ was missing; I inserted an extra line which should help clarify.
A: Start with: $(p)(\frac{1}{2}(2p-1)^n) + (1-p)(-\frac{1}{2}(2p-1)^n) = x$
 (I'm setting it equal to $x$ so it's easier to follow.)  
$\therefore (p)(2p-1)^{n} + (1-p)(-1)(2p-1)^{n} = 2x$
Factor the $(2p-1)^{n}$:
$\therefore (2p-1)^{n} \cdot (p + (1-p)(-1)) = 2x$
$\therefore (2p-1)^{n} \cdot (p + (p-1)) = 2x$
$\therefore (2p-1)^{n} \cdot (2p-1) = 2x$
By law of exponent addition:
$(2p-1)^{n} \cdot (2p-1)^{1} = 2x$
$(2p-1)^{n+1} = 2x$
$\therefore \frac{1}{2} \cdot (2p-1)^{n+1} = x$
And there we are at the solution!
$ = \frac{1}{2} \cdot (2p-1)^{n+1}$
A: Just take $\frac{1}{2}(2p-1)^n$ common and simplify.
$(p)(\frac{1}{2}(2p-1)^n) + (1-p)(-\frac{1}{2}(2p-1)^n)$
Taking $\frac{1}{2}(2p-1)^n$ common, we have
$\frac{1}{2}(2p-1)^n\left(p-(1-p)\right)$
$\frac{1}{2}(2p-1)^n\left(2p-1\right)$
$\frac{1}{2}(2p-1)^{n+1}$
