Equation in natural number

  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. The identity is wrong, it should be

    [tex]
    \sum_{i=0}^{n}2^{n-i} {n+i \choose i}=2^{2 n}
    [/tex]
     
  4. ok my foult. so how can i solve that equation?
     
  5. Well, this
    [tex]
    \sum_{i=0}^{n}2^{n-i} {n+i \choose i}=2^{2 n}
    [/tex]

    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
    [tex]
    \sum_{i=0}^{n}2^{n-i} {n+i \choose i}=2^{2^{n}}
    [/tex]
    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. 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}
    [/tex]
    is good for every natural numbers
     
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