
#1
Jan1111, 11:09 AM

P: 10

How can I show that:
[tex]\sum_{i=0}^{n}2^{ni} {n+i \choose i}=2^{2^{n}} [/tex] for every natural numbers 



#2
Jan1211, 09:32 AM

P: 8

The identity is wrong, it should be
[tex] \sum_{i=0}^{n}2^{ni} {n+i \choose i}=2^{2 n} [/tex] 



#3
Jan1211, 09:54 AM

P: 10

ok my foult. so how can i solve that equation?




#4
Jan1211, 10:45 AM

P: 8

Equation in natural number
Well, this
[tex] \sum_{i=0}^{n}2^{ni} {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^{ni} {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
Jan1211, 11:17 AM

P: 10

How can I show that [tex] \sum_{i=0}^{n}2^{ni} {n+i \choose i}=2^{2 n} [/tex] is good for every natural numbers 


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