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Sums/Series, basic algebra ?

  1. Apr 20, 2008 #1
    [tex]\sum[/tex] (2^(2n)-(-7)^n)/(11^n)

    The book has that expression equal to
    [tex]\sum[/tex] (4/11)^n - [tex]\sum[/tex] (-7/11)^n

    I'm not seeing how the first part changes to (4/11)^n. Wouldn't it be (2^2+2^n) & not 4^n? Or is there something else I'm missing?
     
  2. jcsd
  3. Apr 20, 2008 #2

    nicksauce

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    Fact:
    [tex](x^y)^z = x^{yz}[/tex].
    Does that help you out?
     
  4. Apr 20, 2008 #3
    Crude, nm. 2am & math doesn't mix sometimes. ^_^
     
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