# Math Help - Partial fractions

1. ## Partial fractions

I'm stuck on b) iii) of this question. I'll show you what i have got so far.

the function f is given by f(x)= $\frac{9}{(1+2x)(4 - x)}$

a) express f(x) as a partial fraction.

I got $\frac{2}{(1+2x)} + \frac{1}{(4 -x)}$

b) i) Find the first three terms in the expansion of $\frac{1}{4 -x}$

I got $\frac{1}{4} + \frac{x}{16} + \frac{x^2}{64}$

b) ii) Obtain a similar expansion for $\frac{1}{1+2x}$

I got $1 - 2x + 4x^2$

where i am stuck on is b) iii)

Hence, or otherwise, obtain the first three terms in the expansion of f(x) in ascending powers of x

2. Hello Beard
Originally Posted by Beard
I'm stuck on b) iii) of this question. I'll show you what i have got so far.

the function f is given by f(x)= $\frac{9}{(1+2x)(4 - x)}$

a) express f(x) as a partial fraction.

I got $\frac{2}{(1+2x)} + \frac{1}{(4 -x)}$

b) i) Find the first three terms in the expansion of $\frac{1}{4 -x}$

I got $\frac{1}{4} + \frac{x}{16} + \frac{x^2}{64}$

b) ii) Obtain a similar expansion for $\frac{1}{1+2x}$

I got $1 - 2x + 4x^2$

where i am stuck on is b) iii)

Hence, or otherwise, obtain the first three terms in the expansion of f(x) in ascending powers of x
But you've done all the hard work! - and you've done it right so far. You're going to kick yourself, I think, because all you need to do is to use your Partial Fractions answer:

$f(x) = 2(1+2x)^{-1} + (4-x)^{-1}$

together with your answers to b(i) and b(ii) to write the answer down. Can you see it now?