Equation in natural number

oszust001

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

The identity is wrong, it should be

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

oszust001

ok my foult. so how can i solve that equation?

AtomSeven

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

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