Find the first four terms in the series expansion of (1/(3x+5)) ascending power x and state the set of values of x for which this expansion is valid .


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Question Number 131054 by benjo_mathlover last updated on 01/Feb/21

$Find the first four terms in the series$ $expansion of \frac{1}{3 x + 5} ascending power$ $x and state the set of values of x for$ $which this expansion is valid .$
	Answered by EDWIN88 last updated on 01/Feb/21
	$remark binomial theorem for n∉N and ∣X∣<1 is (1+X)^n =1+nX+((n(n−1))/(2!))X^2 +((n(n−1)(n−2))/(3!))X^3 +... then transform (1/(3x+5)) into form (1+X)^n by writting (1/(3x+5))=(1/(5(1+3x/5)))=(1/5)(1+((3x)/5))^(−1) using n=−1 ,X=((3x)/5) (1/(3x+5))=(1/5){ 1+(−1)((3x)/5)+(((−1)(−2))/(2!))(((3x)/5))^2 +(((−1)(−2)(−3))/(3!))(((3x)/5))^3 +... } = (1/5)−((3x)/(25))+((9x^2 )/(125))−((27x^3 )/(625))+... because the binomial of(1+X)^n valid for ∣X∣<1 then ∣((3x)/5)∣<1 the set of the values of x is {x: −(5/3)<x<(5/3) }$
	$r e m a r k b i n o m i a l t h e o r e m$ $f o r n \notin N a n d ∣ X ∣< 1 i s {(1 + X)}^{n} = 1 + n X + \frac{n (n - 1)}{2!} X^{2} + \frac{n (n - 1) (n - 2)}{3!} X^{3} + . . .$ $t h e n t r a n s f o r m \frac{1}{3 x + 5} i n t o f o r m$ ${(1 + X)}^{n} b y w r i t t i n g \frac{1}{3 x + 5} = \frac{1}{5 (1 + 3 x / 5)} = \frac{1}{5} {(1 + \frac{3 x}{5})}^{- 1}$ $u s i n g n = - 1, X = \frac{3 x}{5}$ $\frac{1}{3 x + 5} = \frac{1}{5} {1 + (- 1) \frac{3 x}{5} + \frac{(- 1) (- 2)}{2!} {(\frac{3 x}{5})}^{2} + \frac{(- 1) (- 2) (- 3)}{3!} {(\frac{3 x}{5})}^{3} + . . .}$ $= \frac{1}{5} - \frac{3 x}{25} + \frac{9 x^{2}}{125} - \frac{27 x^{3}}{625} + . . .$ $b e c a u s e t h e b i n o m i a l o f {(1 + X)}^{n}$ $v a l i d f o r ∣ X ∣< 1 t h e n ∣ \frac{3 x}{5} ∣< 1$ $t h e s e t o f t h e v a l u e s o f x i s {x : - \frac{5}{3} < x < \frac{5}{3}}$
	Answered by mr W last updated on 01/Feb/21

	$\frac{1}{3 x + 5}$ $= \frac{1}{5} \times \frac{1}{1 - (- \frac{3 x}{5})}$ $= \frac{1}{5} [1 + (- \frac{3 x}{5}) + {(- \frac{3 x}{5})}^{2} + {(- \frac{3 x}{5})}^{3} + . . .]$ $= \frac{1}{5} (1 - \frac{3 x}{5} + \frac{9 x^{2}}{25} - \frac{27 x^{3}}{125} + . . .)$ $= \frac{1}{5} - \frac{3 x}{25} + \frac{9 x^{2}}{125} - \frac{27 x^{3}}{625} + . . .$ $w i t h ∣ - \frac{3 x}{5} ∣< 1$ $\Rightarrow - \frac{5}{3} < x < \frac{5}{3}$
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