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

  1. #1
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    Question 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]
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  2. #2
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    Note that $\displaystyle \frac{n-2}{4} = \frac{n}{4} - \frac{1}{2}$.

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


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

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

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