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Equation in natural number

  1. Jan 11, 2011 #1
    How can I show that:
    [tex]\sum_{i=0}^{n}2^{n-i} {n+i \choose i}=2^{2^{n}} [/tex]
    for every natural numbers
  2. jcsd
  3. Jan 12, 2011 #2
    The identity is wrong, it should be

    \sum_{i=0}^{n}2^{n-i} {n+i \choose i}=2^{2 n}
  4. Jan 12, 2011 #3
    ok my foult. so how can i solve that equation?
  5. Jan 12, 2011 #4
    Well, this
    \sum_{i=0}^{n}2^{n-i} {n+i \choose i}=2^{2 n}

    is an identity it is true for all [tex]n[/tex] but, if I understand correctly, you may ask for the values of [tex]n[/tex] that make
    \sum_{i=0}^{n}2^{n-i} {n+i \choose i}=2^{2^{n}}
    true. In this case we have the equation [tex]2n=2^{2^{n}}[/tex], and the solutions are [tex]n \in \lbrace 1,2 \rbrace[/tex].
  6. Jan 12, 2011 #5
    Version of AtomSeven is good.
    How can I show that [tex]
    \sum_{i=0}^{n}2^{n-i} {n+i \choose i}=2^{2 n}
    is good for every natural numbers
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