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# binomial expansion (p3) watch

1. hey

i know how to do the 'usual' type of expansion for a bracket to a power, but how do you do it for a fraction which you have split into partial fractions?

e.g. find the series expansion in ascending powers of x, up to and including the term x cubed for the following: 14/(1-2x) - 6/(2+x)

thanks!
2. (Original post by jazzyj)
hey

i know how to do the 'usual' type of expansion for a bracket to a power, but how do you do it for a fraction which you have split into partial fractions?

e.g. find the series expansion in ascending powers of x, up to and including the term x cubed for the following: 14/(1-2x) - 6/(2+x)

thanks!
Do each part separately, then subtract the second part from the first.
3. you do each fraction seperately, to x³, and add up (or take away, as it might be) the coefficients in the same powers of x.
4. hmm but there's none of the usual (1+x) to a power stuff, so how do i expand each bit?
5. (Original post by jazzyj)
hmm but there's none of the usual (1+x) to a power stuff, so how do i expand each bit?
For the first one, expanding (1+2x), just do as you would with x, but put 2x in each time instead.
For the second, take out a factor 1/2, so that you have 3/(1+x/2)
6. for the first part you just replace the x in the formula with 2x and for the second bit you have to take 2 out so you have [2(1+x/2)]^-1
7. ok thank you...i think ive got it!

xx
8. And don't forget your limits on x if your using the binomial expansion for (1+x)^n where x is not a natural number (0,1,2,3,4,5....). It's always good practice andmay well gain you a mark in the exam.

In this case:

-1<-2x<1 so -0.5<x<0.5 is one limit
-1<-0.5x<1 so -2<x<2x is the other.

So the limits on x would be:
-0.5<x<0.5

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