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Thread: Factoring 2^{^{n-3}}-2^{^{n-2}}

  1. #1

    Factoring 2^{^{n-3}}-2^{^{n-2}}

    Hi,

    So right now I'm studying for an exam and I'm doing a problem on successions.

    The problems asks whether

    ln+1 - ln

    is positive or negative.

    ln = 1/(2^(n-3))

    I was able to solve most of it, but then I got stuck; so I looked up a solution in symbolab:

    https://www.symbolab.com/solver/polynomial-equation-calculator/%5Cfrac%7B1%7D%7B2%5E%7B%5E%7B%5Cleft(n%2B1%5Crigh t)-3%7D%7D%7D-%5Cfrac%7B1%7D%7B2%5E%7B%5E%7Bn-3%7D%7D%7D

    The part that I don't understand about this solution is from here forward (in the nominator):

    2^(^(n-3))-2^(^{n-2))

    where the website factors out the common term 2^(-3)*2^n of

    2^(^(n-3))-2^(^{n-2))


    to get

    -2^(n-3)


    Can someone help me understand this factoring? I'm confused by how the site got to this final answer. How does it factor out
    2^(-3)*2^n ?

    If you can help me understand this I would really appreciate it.

    P.S. Sorry if the formatting is horrible. I'm still figuring out how to use the site



  2. #2
    Elite Member
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    Quote Originally Posted by skatinima View Post
    Hi,
    ln+1 - ln

    is positive or negative.

    ln = 1/(2^(n-3))

    The part that I don't understand about this solution is from here forward (in the nominator):

    2^(^(n-3))-2^(^{n-2))

    where the website factors out the common term 2^(-3)*2^n of

    2^(^(n-3))-2^(^{n-2))


    to get

    -2^(n-3)


    Typically, just piece it together. At first, we're just finding a common denominator.

    [tex]\dfrac{1}{2^{n-3}} - \dfrac{1}{2^{n-2}} = \dfrac{1}{2^{n-3}} - \dfrac{2}{2^{n-3}}[/tex]

    Do you see that this is so? Looks like we're almost done and we haven't done any factoring.
    "Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.

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