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Equation in natural number

by oszust001
Tags: equation, natural, number
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oszust001
#1
Jan11-11, 11:09 AM
P: 10
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
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AtomSeven
#2
Jan12-11, 09:32 AM
P: 8
The identity is wrong, it should be

[tex]
\sum_{i=0}^{n}2^{n-i} {n+i \choose i}=2^{2 n}
[/tex]
oszust001
#3
Jan12-11, 09:54 AM
P: 10
ok my foult. so how can i solve that equation?

AtomSeven
#4
Jan12-11, 10:45 AM
P: 8
Equation in natural number

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
#5
Jan12-11, 11:17 AM
P: 10
Quote Quote by AtomSeven View Post
The identity is wrong, it should be

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


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