F.6 Pure Maths (Binomial thm)

1.

since (nCm)(2^m) = (nCm)(1+1)^m

=(nCm)(mC0+mC1+mC2+mC3+.....+mCm)

=(nCm)(mC0) + (nCm)(mC1) + (nCm)(mC2)+.....+(nCm)(mCm)------(1)

because

(nCm)(mCr) = (n!/m!(n-m)!)(m!/r!(m-r)!)

=n!/(n-m)!(m-r)!r!

=(n!/r!)(1/(n-m)!(m-r)!)

=(n!/(n-r)!r!)( (n-r)!/(m-r)!(n-m)!)

=(nCr)( (n-r)!/(m-r)!(n-r -(m-r))! = (nCr)(n-rCm-r)

therefore (1) will become

(nC0)(nCm) + (nC1)(n-1Cm-1) + (nC2)(n-2Cm-2)+.....+(nCm)(n-mC0) = L.H.S

2) (1-x)^n/x=1/x - nC1 + nC2x^1 - nC3x^2+.........+(-1)^n nCnx^n-1
1/x - (1-x)^n/x = nC1 -nC2x^1 + nC3x^2+.........+(-1)^n-1(nCnx^n-1)
by applying 1-y^n = (1-y)(1+y+y^2+....y^n-1)
we have
(1+(1-x) + (1-x)^2+....+(1-x)^n-1) = nC1 -nC2x^1 + .........+(-1)^n-1(nCnx^n-1)

integrate both sides from x=1 to 0
(x -(1-x)^2/2 -(1-x)^3/3 -....-(1-x)^n/n =nC1x -nC2x^2/2+.....
.....+(-1)^n-1 nCnx^n/n
from x=1 to 0 we have
1 - (-1/2 -1/3 -...-1/n) = nC1-nC2/2+.....+(-1)^n-1(nCn/n)
1+1/2+1/3+....+1/n = nC1-nC2/2+......+(-1)^n-1(nCn/n)

1. Prove that (nCo)(nCm) + (nC1)(n-1Cm-1) + … + (nCm)(n-mCo) = (nCm) (2^m)
, given n>m

2. Prove that (nC1 /1)–(nC2 /2) + (nC3 /3) - … + (-1)^n-1 (nCn / n) = 1 + 1/2 + 1/3 + ... + 1/n


(nCr)(n-rCm-r)
=[n!/r!(n-r)!][(n-r)!/(m-r)!(n-m)!]
=[n!/r!][1/(m-r)!(n-m)!]
=[m!/r!(m-r)!][n!/m!(n-m)!]
=(mCr)(nCm)

(1+x)^m=(mC0)+(mC1)x+ ... +(mCm)x^m
Put x=1,
(mC0)+(mC1)x+ ... +(mCm)x^m=2^m

(nCo)(nCm) + (nC1)(n-1Cm-1) + … + (nCm)(n-mCo)
=(mC0)(nCm)+(mC1)(nCm)+ ... +(mCm)(nCm)
=[(mC0)+(mC1)+ ... +(mCm)](nCm)
=2^m(nCm)

-----------------------------------------------------
d/dx[(nC1 /1)x–(nC2 /2)x^2 + (nC3 /3)x^3 - … + (-1)^n-1 (nCn / n)x^n]
=(nC1)–(nC2)x + (nC3)x^2 - … + (-1)^n-1 (nCn)x^n-1
=-[-(nC1)x + (nC2)x^2 - (nC3)x^3 + … + (-1)^n (nCn)x^n]/x
=[1 -(1-x)^n]/[1-(1-x)]
=1+(1-x)+(1-x)^2+...+(1-x)^n-1

Integrate with upper limit=1, lower limit=0
then result follows

2006-12-06 23:43:24 補充：
呢兩條數花了我不少時間因為我校還沒正式教Binomial…在兩邊使用積分的方面以Definite integral方式會較暢順，比起Indefinite integral更常用最少能避免求出積分常數的問題