Can You Help Prove This Combinatorial Identity?

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Discussion Overview

The discussion centers around proving a specific combinatorial identity involving binomial coefficients. Participants explore various approaches to establish the validity of the identity, which relates to combinatorial proofs and algebraic transformations.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested, Mathematical reasoning

Main Points Raised

  • One participant presents the identity to be proven and shares their initial attempt using factorial transformations, which they found unsuccessful.
  • Another participant applies Pascal's identity to derive the combinatorial identity, suggesting that both algebraic and combinatorial proofs exist for it.
  • A later post introduces a different mathematical expression involving binomial coefficients, but its relevance to the original identity is unclear.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the proof of the identity, and there are competing approaches presented without resolution.

Contextual Notes

The discussion includes assumptions about the validity of Pascal's identity and the conditions under which the original identity holds, but these are not fully explored or resolved.

Who May Find This Useful

Readers interested in combinatorial mathematics, binomial coefficients, and proof techniques may find this discussion relevant.

Lancelot1
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Dear All,

I am trying to prove the following identity:

\[\binom{n}{k}=\binom{n-2}{k}+2\binom{n-2}{k-1}+\binom{n-2}{k-2}\]

My attempt was based on transforming the binomial coefficients into fractions with factorials, and then elimintating similar expressions. Somehow it didn't work out.

I believe that this proof shouldn't be too long. Can you assist ?Thank you in advance.
 
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Using Pascal's identity,
\[
\binom{n}{k}=\binom{n-1}{k-1}+\binom{n-1}{k}=\left[\binom{n-2}{k-2}+\binom{n-2}{k-1}\right]+\left[\binom{n-2}{k-1}+\binom{n-2}{k}\right]=\binom{n-2}{k-2}+2\binom{n-2}{k-1}+\binom{n-2}{k}.
\]
The identity itself has both algebraic and combinatorial proofs.
 
Thank you !
 
Please this:
##2^{p-1}\left(2^{p}-1\right)=\sum_{k=0}^{n=\left\lfloor\frac{p-1}{4}\right\rfloor}\binom{2p+2}{4k+p-4n-1}##
 
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