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: 10 ok my foult. so how can i solve that equation?
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 $$n \in \lbrace 1,2 \rbrace$$.
P: 10
 Quote by AtomSeven 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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