I have expressed the expression as partial fractions and I have done binomial expansion up to x cubed. The answer in the book has different coefficients of x squared and x cubed. I have had a good look at it. Can someone check it?
I have expressed the expression as partial fractions and I have done binomial expansion up to x cubed. The answer in the book has different coefficients of x squared and x cubed. I have had a good look at it. Can someone check it?
Oh right. Sorry, I was using 1/(x-2)
In your simplification line of the first term.
$\displaystyle \dfrac{-1.-2}{2!} \left(\dfrac{x}{4}\right)^2 \neq \dfrac{3x^2}{32}$
$\displaystyle \dfrac{-1.-2}{2!} \left(\dfrac{x}{4}\right)^2 = \dfrac{x^2}{16}$
The binomial coefficient is not to add the numerator but to multiply them.