Given that −1<x<1, find the expansion of ((3−2x)/((1+x)(4+x^2 ))) in ascending power of x, up to and including the term in x^3


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Question Number 161823 by MathsFan last updated on 22/Dec/21

$G i v e n t h a t - 1 < x < 1, f i n d t h e$ $e x p a n s i o n o f \frac{3 - 2 x}{(1 + x) (4 + x^{2})} i n$ $a s c e n d i n g p o w e r o f x, u p t o a n d$ $i n c l u d i n g t h e t e r m i n x^{3}$
	Answered by mr W last updated on 23/Dec/21

	$= \frac{3 (1 - \frac{2 x}{3})}{4 (1 + x) (1 + \frac{x^{2}}{4})}$ $= \frac{3}{4} (1 - \frac{2 x}{3}) (1 - x + x^{2} - x^{3} + x^{4} - . . .) (1 - \frac{x^{2}}{4} + \frac{x^{4}}{16} - . . .)$ $= \frac{3}{4} [1 + (- \frac{2}{3} - 1) x + (1 - \frac{1}{4} + \frac{2}{3}) x^{2} + (- 1 - \frac{2}{3} + \frac{2}{3 \times 4} + \frac{1}{4}) x^{3} + . . .]$ $= \frac{3}{4} [1 - \frac{5}{3} x + \frac{17}{12} x^{2} - \frac{5}{4} x^{3} + . . .]$ $= \frac{3}{4} - \frac{5}{4} x + \frac{17}{16} x^{2} - \frac{15}{16} x^{3} + . . .$
	Commented by peter frank last updated on 23/Dec/21

	$thank you$
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