Equation in natural number

  • Thread starter oszust001
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  • #1
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Main Question or Discussion Point

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
 

Answers and Replies

  • #2
8
0
The identity is wrong, it should be

[tex]
\sum_{i=0}^{n}2^{n-i} {n+i \choose i}=2^{2 n}
[/tex]
 
  • #3
10
0
ok my foult. so how can i solve that equation?
 
  • #4
8
0
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].
 
  • #5
10
0
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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