Is This the Correct Identity for Natural Numbers?

  • Context: Graduate 
  • Thread starter Thread starter oszust001
  • Start date Start date
  • Tags Tags
    Natural
Click For Summary

Discussion Overview

The discussion revolves around the identity involving the summation of binomial coefficients and powers of two, specifically whether the identity \(\sum_{i=0}^{n}2^{n-i} {n+i \choose i}=2^{2^{n}}\) holds for all natural numbers, and the alternative identity \(\sum_{i=0}^{n}2^{n-i} {n+i \choose i}=2^{2 n}\). Participants are exploring the validity of these identities and seeking methods to demonstrate their correctness.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions how to show that \(\sum_{i=0}^{n}2^{n-i} {n+i \choose i}=2^{2^{n}}\) holds for every natural number.
  • Another participant asserts that the identity is incorrect and proposes that it should be \(\sum_{i=0}^{n}2^{n-i} {n+i \choose i}=2^{2 n}\).
  • A participant acknowledges their mistake and seeks guidance on solving the equation.
  • One participant claims that the identity \(\sum_{i=0}^{n}2^{n-i} {n+i \choose i}=2^{2 n}\) is true for all \(n\) and discusses the conditions under which \(\sum_{i=0}^{n}2^{n-i} {n+i \choose i}=2^{2^{n}}\) might hold, suggesting that the equation \(2n=2^{2^{n}}\) has solutions \(n \in \{1,2\}\).
  • Another participant reiterates the claim that the identity is incorrect and supports the assertion that \(\sum_{i=0}^{n}2^{n-i} {n+i \choose i}=2^{2 n}\) is valid for all natural numbers.

Areas of Agreement / Disagreement

Participants express disagreement regarding the correctness of the original identity, with some asserting it is incorrect and others defending its validity under certain conditions. The discussion remains unresolved regarding the general applicability of the identities.

Contextual Notes

Participants have not reached a consensus on the validity of the identities, and there are unresolved assumptions regarding the conditions under which each identity may hold true.

oszust001
Messages
10
Reaction score
0
How can I show that:
\sum_{i=0}^{n}2^{n-i} {n+i \choose i}=2^{2^{n}}
for every natural numbers
 
Physics news on Phys.org
The identity is wrong, it should be

<br /> \sum_{i=0}^{n}2^{n-i} {n+i \choose i}=2^{2 n} <br />
 
ok my foult. so how can i solve that equation?
 
Well, this
<br /> \sum_{i=0}^{n}2^{n-i} {n+i \choose i}=2^{2 n} <br />

is an identity it is true for all n but, if I understand correctly, you may ask for the values of n that make
<br /> \sum_{i=0}^{n}2^{n-i} {n+i \choose i}=2^{2^{n}} <br />
true. In this case we have the equation 2n=2^{2^{n}}, and the solutions are n \in \lbrace 1,2 \rbrace.
 
AtomSeven said:
The identity is wrong, it should be

<br /> \sum_{i=0}^{n}2^{n-i} {n+i \choose i}=2^{2 n} <br />
Version of AtomSeven is good.
How can I show that <br /> \sum_{i=0}^{n}2^{n-i} {n+i \choose i}=2^{2 n} <br />
is good for every natural numbers
 

Similar threads

Graduate Coefficients of Chebyshev polynomials
  • · Replies 2 ·
Replies
2
Views
3K
Undergrad How to Prove a Congruence Relation for Prime Numbers?
  • · Replies 17 ·
Replies
17
Views
2K
Undergrad P-adic valuation expression for a given natural number
  • · Replies 9 ·
Replies
9
Views
2K
Undergrad ##S_{n}(\alpha)=\sum_{k=1}^{n} (-1)^{\lfloor{k\alpha\rfloor}}##
  • · Replies 2 ·
Replies
2
Views
2K
Undergrad Proving inequality about dimension of quotient vector spaces
  • · Replies 25 ·
Replies
25
Views
4K
Undergrad Confused about small detail in rank-nullity theorem
  • · Replies 1 ·
Replies
1
Views
3K
Undergrad Number of Binary Operations on a Set with a Special Property
  • · Replies 2 ·
Replies
2
Views
3K
Graduate Completeness of the formal power series and valued fields
  • · Replies 2 ·
Replies
2
Views
2K
Undergrad Dfns group action & group, written in terms of the other
  • · Replies 26 ·
Replies
26
Views
1K
Undergrad Convergence and divergence of series and sequences
  • · Replies 7 ·
Replies
7
Views
3K