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I'm stuck in the follow equation: $\dfrac{2^{n + 4} + 2^{n + 2} + 2^{n - 1}}{2^{n - 2} + 2^{n - 1}}$

As all the bases are equal, I got $\dfrac{3n + 5}{2n - 3}$

Where I've to go now ?

Thanks

EDIT:

Then, my initial idea was totally wrong, starting again, in the right way I got:

$\dfrac{2^n(2^4 + 2^2 + 2^{-1})}{2^n(2^{-2} + 2^{-1})}$

but it's still wrong, I didn't get the right idea on the divisions you have shown to me in the answers.

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    @aajjbb: What you added is perfectly correct. Now cancel the $2^n$. But in order to get something that "looks nice" (integer divided by integer) it is helpful to multiply your new top and bottom by $4$ (that is, $2^2$).2012-06-22

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You don't have an equation, you are probably asked to simplify. Divide top and bottom by $2^{n-2}$.

When you do that, at the bottom you will have $1+2$. On top you will have $2^6+2^4+2$.

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    There are several ways to think of it. You can say that by the rules of exponentiation, $2^{n+4}=2^{n-2}2^6$. Now we can cancel $2^{n-2}$ on top and bottom. Or else we can view $2^{n+4}$ as the product of $n+4$ $2$'s, and $2^{n-2}$ as the product of $n-2$ $2$'s, which is $6$ fewer. So when we divide we get the product of $6$ $2$'s left, that is, $2^6$. However, I think that the "factoring" explanation of Brian M. Scott may be more directly suited to your needs.2012-06-22
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You’ve nothing to solve. My guess is that you’re supposed to simplify the fraction:

$\begin{align*} \dfrac{2^{n + 4} + 2^{n + 2} + 2^{n - 1}}{2^{n - 2} + 2^{n - 1}}&=\frac{2^{n-1}(2^5+2^3+1)}{2^{n-2}(1+2)}\\ &=\frac{2^{n-1}}{2^{n-2}}\cdot\frac{32+8+1}{3}\;, \end{align*}$

and you should have no trouble finishing it.

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    @talmid: Thanks. (At least I was consistent after the typo!)2012-06-22