Simplifying $\frac{2^{n + 4} + 2^{n + 2} + 2^{n - 1}}{2^{n - 2} + 2^{n - 1}}$

Question

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.

Answer 1

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 $2^6+2^4+2$.

Answer 2

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.