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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:
$$\sum_{i=0}^{n}2^{n-i} {n+i \choose i}=2^{2^{n}}$$
for every natural numbers

 AtomSeven Jan12-11 09:32 AM

Re: Equation in natural number

The identity is wrong, it should be

$$\sum_{i=0}^{n}2^{n-i} {n+i \choose i}=2^{2 n}$$

 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
$$\sum_{i=0}^{n}2^{n-i} {n+i \choose i}=2^{2 n}$$

is an identity it is true for all $$n$$ but, if I understand correctly, you may ask for the values of $$n$$ that make
$$\sum_{i=0}^{n}2^{n-i} {n+i \choose i}=2^{2^{n}}$$
true. In this case we have the equation $$2n=2^{2^{n}}$$, and the solutions are $$n \in \lbrace 1,2 \rbrace$$.

 oszust001 Jan12-11 11:17 AM

Re: Equation in natural number

Quote:
 Quote by AtomSeven (Post 3078979) The identity is wrong, it should be $$\sum_{i=0}^{n}2^{n-i} {n+i \choose i}=2^{2 n}$$
Version of AtomSeven is good.
How can I show that $$\sum_{i=0}^{n}2^{n-i} {n+i \choose i}=2^{2 n}$$
is good for every natural numbers

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