1. ## Algebra with exponents

Hi all-

I need to see the steps to get

(1- 1/2 + 1/2*e^t)^25 * (1- 1/2 + 1/2*e^(-t))^25 * e^(25t)

to be (1- 1/2 + 1/2*e^t)^50

Can anyone help, i am having trouble seeing how?

Brian

2. Originally Posted by B_Miner
Hi all-

I need to see the steps to get

(1- 1/2 + 1/2*e^t)^25 * (1- 1/2 + 1/2*e^(-t))^25 * e^(25t)

to be (1- 1/2 + 1/2*e^t)^50

Can anyone help, i am having trouble seeing how?

Brian
Probably there is a typo in your question because 1 - 1/2 = 1/2 . I don't understand why you didn't simplify this sum first.

Nevertheless I show you what to do:

$\displaystyle \left(1-\frac12 + \frac12 \cdot e^t\right)^{25} \cdot \left(1-\frac12 + \frac12 \cdot e^{-t}\right)^{25} \cdot e^{25t} = \left(\frac12 + \frac12 \cdot e^t\right)^{25} \cdot \left(e^t \left(\frac12 + \frac12 \cdot e^{-t}\right) \right)^{25}$ $\displaystyle = \left(\frac12 + \frac12 \cdot e^t\right)^{25} \cdot \left(\frac12 \cdot e^t+ \frac12 \right)^{25} = \left(\frac12 + \frac12 \cdot e^t\right)^{50}$

3. Thanks very much. I struggle trying to "see the end" and making the right moves.

PS the reason it was not simplified is that the result should be the mgf of the binomial random variable.

4. Originally Posted by B_Miner
[snip]the result should be the mgf of the binomial random variable.
http://www.mathhelpforum.com/math-he...tribution.html