Binomial Coefficients Identity

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SUMMARY

The forum discussion focuses on proving the binomial coefficients identity for integers n greater than or equal to 2, specifically the expression nC1 - 2nC2 + 3nC3 - ... = 0. Additionally, it addresses the identity 2x1 nC2 + 3x2 nC3 + 4x3 nC4 + ... = n(n-1)2^(n-2). Participants reference the binomial theorem, specifically (1+t)^a = 1 + aC1(t) + aC2(t^2) + ..., to derive these identities and suggest using derivatives of the binomial expansion for further insights.

PREREQUISITES
  • Understanding of binomial coefficients (nCm)
  • Familiarity with the binomial theorem
  • Basic knowledge of factorials
  • Experience with derivatives in calculus
NEXT STEPS
  • Explore the application of the binomial theorem in combinatorial proofs
  • Learn how to derive identities using derivatives of (1+x)^n
  • Investigate properties of alternating sums of binomial coefficients
  • Study advanced combinatorial identities and their proofs
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Mathematics students, educators, and anyone interested in combinatorial identities and proofs will benefit from this discussion.

murmillo
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Homework Statement


Prove that for an integer n greater than or equal to 2,

nC1 - 2nC2 + 3nC3 - + ... = 0. (nCm means n choose m)

Also,
2x1 nC2 + 3x2 nC3 + 4x3 nC4 +... = n(n-1)2^(n-2)

Homework Equations



(1+t)^a = 1 + aC1(t) + aC2(t^2) + ...

The Attempt at a Solution


I don't know if these identities will help, but I've found
nC0 - nC1 + nC2 - nC3 + - ... = 0
and
nC0 + nC1 + nC2 +... = 2^n

I tried writing out the given expression in terms of factorials and got
1/0! n - 1/1! n(n-1) + 1/2! n(n-1)(n-2) - 1/3! n(n-1)(n-2)(n-3) + - ...,
but I don't think this is going anywhere.
 
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The binomial theorem is your friend. (1+x)^n=nC0+nC1*x+nC2*x^2+...+nCn*x^n, right? Think about what you might want to put x equal to, and think about taking derivatives of (1+x)^n.
 

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