Equation in natural number

[oszust001]

How can I show that:
\sum_{i=0}^{n}2^{n-i} {n+i \choose i}=2^{2^{n}}
for every natural numbers

[AtomSeven]

The identity is wrong, it should be


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

[oszust001]

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

[AtomSeven]

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]

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