# Math Help - 2^[(n-2)/4] + 2^[(n-4)/4] + 1 I can't get it :(

1. ## 2^[(n-2)/4] + 2^[(n-4)/4] + 1 I can't get it :(

How do I factor out the 2 so I can have the same bases?

Can someone take me through this problem?

I think the answer in the end will be 2^[n/4]

2. Note that $\frac{n-2}{4} = \frac{n}{4} - \frac{1}{2}$.

Note that $\frac{n-4}{4} = \frac{n}{4} - 1$.

Therefore $\displaystyle{2^{\frac{n-2}{4}} + 2^{\frac{n-4}{4}}=2^{\frac{n}{4} - \frac{1}{2}} + 2^{\frac{n}{4}-1}}$

$\displaystyle{=2^{\frac{n}{4}}\cdot 2^{-\frac{1}{2}} + 2^{\frac{n}{4}}\cdot 2^{-1}}$

$\displaystyle{=2^{\frac{n}{4}}(2^{-\frac{1}{2}} + 2^{-1})}$