Binomial Theorem Help (1 Viewer)

[Andy005]

Can you please explain the theory of the steps involved in solving this question, as I'm struggling with these types of questions. Also, any tips on proof questions relating to binomial theorem would be much appreciated.

Thanks for your help.

 

[pikachu975]

So it says both sides but to get the second side try simplify what they gave.

[(1+x)(1-x)]^n = (1 - x^2)^n

Now expand left hand side:
(nC0 + nC1 x + ... + nCn x^n)(nC0 - nC1 x + nC2 x^2 - ... - nCn x^n)
Note that the second brackets has nCn x^n as negative as n is odd.
= [(nC0)^2 - nC0.nC1 x + nC0.nC2 x^2 - .... - nC0.nCn x^n] + [nC1.nC0 x - (nC1)^2 x^2 + ... - nC1.nCn x^(n+1)] + ....
From subbing in x = 1 and expanding you notice every term will cancel out except for the squares of the coefficients, i.e. = (nC0)^2 - (nC1)^2 + ... + (nCn)^2

Expand right hand side and sub in x = 1:
nC0 - nC1 + nC2 - ... - nCn

Since they're the same equation equate both sides:
sigma(k=0 to n) (-1)^k (nCk)^2 = nC0 - nC1 + nC2 - .... - nCn
Via symmetry nC0 = nCn and so on so the right hand side equals 0 if you keep using symmetry and you have your answer

EDIT: Fixed mistake

 

[Andy005]

how does (nC0).(nC2)x^2 cancel out?

 

[pikachu975]

how does (nC0).(nC2)x^2 cancel out?

I think I laid it out wrong. There are multiple terms when expanding, so consider the terms when each term in the LHS is expanded:

nC0: nC0.nC2 x^2, - nC0.nC(n-2) x^(2n-2)
nC2: nC2.nC0 x^2, - nC2.nCn x^(n+2)
nCn: nCn.nC2 x^2, - nCn.nC(n-2) x^(2n-2)

Via symmetry these all have a coefficient of nC0.nC2 when subbing x = 1 so they cancel out

My bad it should cancel out after subbing in x = 0, thanks for pointing it out!