nCr ...好難...thx=]

2007-10-27 8:23 pm
Prove that nC(r+1)+2(nCr)+nC(r-1) = (n+2)C(r+1)

回答 (1)

2007-10-27 8:38 pm
✔ 最佳答案
LHS = nC(r+1)+2(nCr)+nC(r- 1)
=n!/(r+1)!(n-r-1)! + 2n!/r!(n-r)! + n!/(r-1)!(n-r+1)!
= [n!/(r-1)!(n-r-1)!] {1/r(r+1) + 2/r(n-r) + 1/(n-r)(n-r+1)}
=[n!/(r-1)!(n-r-1)!]{ (n-r)(n-r+1) + 2(r+1)(n-r+1) +r(r+1)}/r(r+1)(n-r)(n-r+1)
=[n!/(r+1)!(n-r+1)!]{ (n-r)(n-r+1) + 2(r+1)(n-r+1) +r(r+1)}
=[n!/(r+1)!(n-r+1)!]{n^2-2nr+r^2 +n-r + 2nr-2r^2+2r+2n-2r+2 +r^2 + r}
=[n!/(r+1)!(n-r+1)!]{n^2 +3n+2}
=n!(n+1)(n+2)/(r+1)!(n-r+1)!
=(n+2)!/(r+1)!(n-r+1)!
= (n+2)C(r+1)


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