1. ## series

(a)Expand each of the following functions as a series in ascending power of x up to and including the term in x^3 .

(1) $\displaystyle \frac{x-1}{x+1}$

(2) $\displaystyle \frac{1}{(x-2)(1+2x)}$

2. Hello, thereddevils!

(a) Expand each of the following functions as a series
in ascending power of $\displaystyle x$ up to and including the term in $\displaystyle x^3$

$\displaystyle (1)\;\;\frac{x-1}{x+1}$
Have you tried Long Division?

We have: .$\displaystyle \frac{x-1}{x+1} \;=\;\frac{-1+x}{1+x}$

. . $\displaystyle \begin{array}{ccccccccc} & & & -1 & +2x &- 2x^2 & +2x^3 & + \hdots \\ & & -- & -- & -- & -- & -- & -- \\ 1+x & ) & -1 & +x \\ & & -1 & -x \\ & & -- & -- \\ & & & 2x \\ & & & 2x & +2x^2 \\ & & & -- & -- \\ & & & & -2x^2 \\ & & & & -2x^2 & -2x^3 \\ \end{array}$

. . . . . . . . . . . . . . . $\displaystyle \begin{array}{ccccc} -- & \;\;-- \\ & \;\; 2x^3 \\ &\;\; 2x^3 &\;\; + 2x^4 \\ &\;\; -- & \;\;-- \\ & & \;\;\hdots \end{array}$

Therefore: .$\displaystyle \frac{x-1}{x+1} \;=\;-1 + 2x - 2x^2 + 2x^3 - \hdots$

3. Thanks Soroban . So if we want to swap positions , we hv to swap the top and bottom as well right ? Is this the commutative law ?

Wow , i have never realised we can do it by long division but can we do it using binomial theorem ?