Show That NC0+nC2+nC4+...=nC1+nC3+...=2^,where ^=n-1. - Algebra

Trang chủ » Chứng Minh (nc0)^2 » Show That NC0+nC2+nC4+...=nC1+nC3+...=2^,where ^=n-1. - Algebra

Có thể bạn quan tâm

SOLUTION: Show that nC0+nC2+nC4+...=nC1+nC3+...=2^,where ^=n-1. Algebra -> Permutations -> SOLUTION: Show that nC0+nC2+nC4+...=nC1+nC3+...=2^,where ^=n-1. Log On
Algebra: Combinatorics and PermutationsSection SolversSolvers LessonsLessons Answers archiveAnswers
  • Click here to see ALL problems on Permutations
Question 97167: Show that nC0+nC2+nC4+...=nC1+nC3+...=2^,where ^=n-1. Answer by mathslover(157) About Me (Show Source): You can put this solution on YOUR website! we have +%28+1+%2B+x+%29%5E+n = nC0 + nC1x + nC2x^2 + nC3x^3 + ...) Sustituting x= -1 we have 0%5En = nC0 - nC1 + nC2 - nC3 + ... 0 = nC0 - nC1 + nC2 - nC3 + ... grouping all the negative terms on the other side we have nC0 + nC2 + nC4 + .... = nC1 + nC3 + nC5 +.... which completes the first portion of the proof . to prove that each of these expressions evaluate to 2%5E%28n-1%29 Substitute x =1 in the expansion of +%28+1+%2B+x+%29%5E+n we have %282%5En%29 = nC0 + nC1 + nC2 + nC3 + ....... %282%5En%29 = (nC0 + nC2 + nC4 +...) + (nC1 + nC3 + nC5 +...) grouping the odd and even terms together Since (nC0 + nC2 + nC4 +...) = (nC1 + nC3 + nC5 +.... ) as proved already we can rewrite the expression above as %282%5En%29 = (nC0 + nC2 + nC4 +...) + (nC0 + nC2 + nC4 +...) 2(nC0 + nC2 + nC4 +...) = %282%5En%29 Dividing by 2 on both sides (nC0 + nC2 + nC4 +...) = %282%5En%29%2F2 = 2%5E%28n-1%29 therefore we have (nC0 + nC2 + nC4 +...) = (nC1 + nC3 + nC5 +.... ) = 2%5E%28n-1%29

Từ khóa » Chứng Minh (nc0)^2