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-   -   Equation in natural number (http://www.physicsforums.com/showthread.php?t=462649)

oszust001 Jan11-11 11:09 AM

Equation in natural number
 
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

AtomSeven Jan12-11 09:32 AM

Re: Equation in natural number
 
The identity is wrong, it should be

[tex]
\sum_{i=0}^{n}2^{n-i} {n+i \choose i}=2^{2 n}
[/tex]

oszust001 Jan12-11 09:54 AM

Re: Equation in natural number
 
ok my foult. so how can i solve that equation?

AtomSeven Jan12-11 10:45 AM

Re: Equation in natural number
 
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].

oszust001 Jan12-11 11:17 AM

Re: Equation in natural number
 
Quote:

Quote by AtomSeven (Post 3078979)
The identity is wrong, it should be

[tex]
\sum_{i=0}^{n}2^{n-i} {n+i \choose i}=2^{2 n}
[/tex]

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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