Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    [tex]
    \sum_{i=0}^{n}2^{n-i} {n+i \choose i}=2^{2 n}
    [/tex]
     
  4. Jan 12, 2011 #3
    ok my foult. so how can i solve that equation?
     
  5. Jan 12, 2011 #4
    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. 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}
    [/tex]
    is good for every natural numbers
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Similar Discussions: Equation in natural number
  1. NCk is a natural number (Replies: 18)

  2. Natural numbers Z? (Replies: 6)

  3. Natural Numbers Proof? (Replies: 5)

Loading...