# Equation in natural number

by oszust001
Tags: equation, natural, number
 P: 10 How can I show that: $$\sum_{i=0}^{n}2^{n-i} {n+i \choose i}=2^{2^{n}}$$ for every natural numbers
 P: 8 The identity is wrong, it should be $$\sum_{i=0}^{n}2^{n-i} {n+i \choose i}=2^{2 n}$$
 P: 8 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 $$n \in \lbrace 1,2 \rbrace$$.
 Quote by AtomSeven The identity is wrong, it should be $$\sum_{i=0}^{n}2^{n-i} {n+i \choose i}=2^{2 n}$$
How can I show that $$\sum_{i=0}^{n}2^{n-i} {n+i \choose i}=2^{2 n}$$