Binomial Expansion: Find (n+1)Ck in Terms of nCj

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    Binomial Expansion
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Discussion Overview

The discussion revolves around finding an expression for the binomial coefficient (n+1)Ck in terms of the coefficients nCj, as derived from the binomial expansion of (a+b)^n. Participants explore various approaches and special cases, including k=0 and k=n+1.

Discussion Character

  • Mathematical reasoning
  • Technical explanation
  • Homework-related

Main Points Raised

  • One participant presents the formula for (n+1)Ck and relates it to nCk using factorials.
  • Another participant expresses doubt about the initial response, indicating that the question has multiple parts and requires further elaboration.
  • A later post references Pascal's triangle, suggesting a relationship where (n+1)Ck can be expressed as the sum of two other binomial coefficients, nC(j-1) and nCj, for k not equal to 0.

Areas of Agreement / Disagreement

Participants do not appear to reach a consensus, as there are differing interpretations and approaches to the problem, with some participants questioning the validity of earlier claims.

Contextual Notes

Some assumptions about the values of k and n are implied but not explicitly stated. The relationship between the coefficients is not fully resolved, and the discussion includes various interpretations of the problem.

saadsarfraz
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Hi, I've been struggling with this problem for sometime. Let nCk be the kth coefficient in the binomial expansion of (a+b)^n. Find an expression for (n+1)Ck in term of the various nCj. Feel free to treat k=0 and k=n+1 as special cases.
 
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(n+1)Ck = (n+1)!/[(n+1-k)!*k!]

vs.

nCk = n!/[(n-k)!k!]

Now (n+1)! = n!*(n+1) and (n+1-k)! = (n+1-k)(n-k)!
 
thanks for the reply but i don't think that's the answer. The question I posted is the 1st part. The second part states: Using the expression you found, (which is the question i posted) show that for all n>=0 and for all k, 0=<k<=n, nCk = n!/(k!)(n-k)!
 
saadsarfraz said:
Hi, I've been struggling with this problem for sometime. Let nCk be the kth coefficient in the binomial expansion of (a+b)^n. Find an expression for (n+1)Ck in term of the various nCj. Feel free to treat k=0 and k=n+1 as special cases.

Pascal's triangle: (n+1)Ck= nC(j-1)+ nCj for k not 0.
 

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