Simplify $(p)(\frac{1}{2}(2p-1)^n) + (1-p)(-\frac{1}{2}(2p-1)^n)$ to $\frac{1}{2}(2p-1)^{n+1}$

Question

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.

You are probably having a minus sign problem. Makes you a member of my club. — André Nicolas, May 09 '13 at 02:06
The answers show me I indeed had that problem. I kinda feel bad for making this question. — user1002327, May 09 '13 at 02:08
Um, I guess you're right. Now I know I have to be more careful with signs. I was cancelling out both $p$. — user1002327, May 09 '13 at 02:46

score 4 · Answer 1 · answered May 09 '13 at 02:07

4

$$p\cdot (1/2)(2p-1)^n +(1-p) \cdot (-1/2)(2p-1)^n = \\ p \cdot (1/2)(2p-1)^n +(p-1) \cdot(1/2)(2p-1)^{\color{red}{n}}= \\ \left[\frac{p}{2}+\frac{p}{2}-\frac{1}{2}\right](2p-1)^n =\\ (2p-1)\cdot(1/2) (2p-1)^n$$

You are correct that an $n$ was missing; I inserted an extra line which should help clarify.

answered May 09 '13 at 02:07

vadim123

82,796

Thank you for the clarification. I'll accept Maazul's (since he answered first). I already upvoted yours. – user1002327 May 09 '13 at 02:12

score 4 · Answer 2 · answered May 09 '13 at 02:13

4

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}$

answered May 09 '13 at 02:13

xisk

812
7
17

Your answer is very detailed too, and I feel embarrassed for missing that $(-1)$. Have an upvote. – user1002327 May 09 '13 at 02:20
Don't feel embarrassed! Learning is good. :-) – xisk May 09 '13 at 02:22

score 3 · Accepted Answer · answered May 09 '13 at 02:04

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

$\frac{1}{2}(2p-1)^n\left(p-(1-p)\right)$

$\frac{1}{2}(2p-1)^n\left(2p-1\right)$

$\frac{1}{2}(2p-1)^{n+1}$

Simplify $(p)(\frac{1}{2}(2p-1)^n) + (1-p)(-\frac{1}{2}(2p-1)^n)$ to $\frac{1}{2}(2p-1)^{n+1}$

3 Answers3