Is This the Correct Identity for Natural Numbers?

[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