Question

Binomial Theorem Question - The first three therms in the binomial expansion of (a+b)^n, in ascending powers of b, are denoted by p, q and r respectively. Show that (q^2/pr)=(2n/2n-1).

Answer 1

I think the first thing to do would be to how the first 3 terms
\[=a ^{n}+\left(\begin{matrix}n \\ 1\end{matrix}\right)a ^{n-1}b+\left(\begin{matrix}n \\ 2\end{matrix}\right)a ^{n-2}b ^{2}\]

Answer 2

Yeah, that's right..I've written out this step already but am stumped after this...

Answer 3

so now we have the terms we should replace them so that
p=a^n
q=nC1*a^(n-1)b
r=nC2*a^(n-2)b^2

Answer 4

we can then put them into q^2/pr
[nC1*a^(n-1)b]^2/a^n*nC2*a^(n-2)b^2

Answer 5

since nC1=n
we can expand it to be
n^2*a^(2n-2)b^2/a^n*nC2*a(n-2)b^2
n^2*a^(2n-2)b^2/nC2*a(2n-2)b^2
n^2/nC2

Answer 6

I don't understand why the a^n in the first line isn't present in the second step..

Answer 7

Oh ok, now I get it..it's multiplied with a^(n-2)

Answer 8

Thanks a LOT!