Binomial Theorem (1 Viewer)

lychnobity · Aug 14, 2009

So I seem to have had a brain malfunction:

Use the binomial theorem to expand (1+x)^p+q and hence write down the term of (1+x)^p+q/x^q which is independent of x

Drongoski · Aug 15, 2009

If I'm not mistaken:

$\100dpi (1 + x)^{p+q}\ = \sum_{k=0}^{p+q}\binom{p+q}{k}x^{k}\\ \\ \therefore \ \ \frac{(1+x)^{p+q}}{x^{q}}\ =\ \sum_{k=0}^{p+q}\binom{p+q}{k}\ x^{k-q}\\ \\ Independent\ of\ x\ means\ the \ power\ of\ x: \ \ \ k-q = 0 \ \Rightarrow \ k=q\\ \\ This\ term\ is\ \ \ \binom{p+q}{q}$

jet · Aug 15, 2009

Yeah, he's right. I remember this question- this was my HSC. It's so long ago now...

Michaelmoo · Aug 15, 2009

NUther way to think about it is get the binomial term that contains x^q, so that when you divide by x^q, your term is independant of x. i.e.

(p+q)Cq

jet · Aug 15, 2009

Michaelmoo said:
NUther way to think about it is get the binomial term that contains x^q, so that when you divide by x^q, your term is independant of x. i.e.

(p+q)Cq

Yeah, but Drongoski is expressing that mathematically.

Binomial Theorem (1 Viewer)

lychnobity

Active Member

Drongoski

Well-Known Member

jet

Banned

Michaelmoo

cbff...

jet

Banned

Users Who Are Viewing This Thread (Users: 0, Guests: 1)