binomial expansion question (1 Viewer)

[smallcattle]

Consider binomial expansion (1+x)^n

1. prove that nC0 + nC1 + nC2 + nC3 +..... + nCn = 2^n

2. prove that nC1 + nC3 + nC5 +... = 2^(n-1)

3. from (1+x)^n(1+x)^n = (1+x)^2n and comparing the coefficient of x^(n+1) prove (nC0)(nC1) + (nC1)(nC2) + (nC2)(nC3) + .... + (nC(n-1))(nCn) = 2n! / (n-1)!(n+1)!

thx

 

[Supra]

expand (1+x)ⁿ then sub in 1 on both sides and u get hte first answer

 

[gordo]

if n is odd, then every second term will be negative

 

[Archman]

expand (1-x)^n and sub in one, and move all the negative terms to one side.
so u have sum of nCeven numbers = sum nCodd numbers,
sub back into part 1 and u'll get part 2

part 3:
try look at the coefficient of the term x^(n+1) on both sides.

 

[nit]

Yeh, or just use (1+x)^n and sub x=-1 as the q asked. Either way is good.

 

[AGB]

If you wanted the answer to part 3, i have typed it out below:

1. expansion of (x+1)^n.(x+1)^n, and letting x=1:

(nC0 + nC1 + nc2 + ... + nCn)(nC0 + nC1 + nc2 + ... + nCn)

2. sum of coefficients of the x^(n+1) terms is:

(nC1)(nCn) + (nC2)(nCn-1) + (nC3)(nCn-2) + ... + (nCn)(nCn-1)

= (nC1)(nC0) + (nC2)(nC1) + (nC3)(nC2) + ... + (nCn)(nCn-1)

3. expansion of (1 + x)^2n is:

2nC0 + 2nC1(x) + 2nC2(x^2) + ... + 2nCn(x^n) + 2nC(n+1).(x^(n+1)) +...

4. coefficient of the x^(n+1) term:

(2n)C(n+1) = (2n)!/(2n - n - 1)!(n + 1)!
= (2n)!/(n - 1)! (n + 1)!

:. (nC0)(nC1) + (nC1)(nC2) + (nC2)(nC3) + .... + (nC(n-1))(nCn) = (2n)! / (n-1)!(n+1)! as required