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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 (1/(3x+5)) ascending power x and state the set of values of x for which this expansion is valid .
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{first}\:\mathrm{four}\:\mathrm{terms}\:\mathrm{in}\:\mathrm{the}\:\mathrm{series}\: \\ $$$$\mathrm{expansion}\:\mathrm{of}\:\frac{\mathrm{1}}{\mathrm{3x}+\mathrm{5}}\:\mathrm{ascending}\:\mathrm{power} \\ $$$$\mathrm{x}\:\mathrm{and}\:\mathrm{state}\:\mathrm{the}\:\mathrm{set}\:\mathrm{of}\:\mathrm{values}\:\mathrm{of}\:\mathrm{x}\:\mathrm{for} \\ $$$$\mathrm{which}\:\mathrm{this}\:\mathrm{expansion}\:\mathrm{is}\:\mathrm{valid}\:. \\ $$
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) }
$${remark}\:{binomial}\:{theorem} \\ $$$${for}\:{n}\notin\mathbb{N}\:{and}\:\mid{X}\mid<\mathrm{1}\:{is}\:\left(\mathrm{1}+{X}\right)^{{n}} =\mathrm{1}+{nX}+\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}!}{X}^{\mathrm{2}} +\frac{{n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)}{\mathrm{3}!}{X}^{\mathrm{3}} +… \\ $$$${then}\:{transform}\:\frac{\mathrm{1}}{\mathrm{3}{x}+\mathrm{5}}\:{into}\:{form} \\ $$$$\left(\mathrm{1}+\mathrm{X}\right)^{{n}} \:{by}\:{writting}\:\frac{\mathrm{1}}{\mathrm{3}{x}+\mathrm{5}}=\frac{\mathrm{1}}{\mathrm{5}\left(\mathrm{1}+\mathrm{3}{x}/\mathrm{5}\right)}=\frac{\mathrm{1}}{\mathrm{5}}\left(\mathrm{1}+\frac{\mathrm{3}{x}}{\mathrm{5}}\right)^{−\mathrm{1}} \\ $$$${using}\:{n}=−\mathrm{1}\:,{X}=\frac{\mathrm{3}{x}}{\mathrm{5}} \\ $$$$\frac{\mathrm{1}}{\mathrm{3}{x}+\mathrm{5}}=\frac{\mathrm{1}}{\mathrm{5}}\left\{\:\mathrm{1}+\left(−\mathrm{1}\right)\frac{\mathrm{3}{x}}{\mathrm{5}}+\frac{\left(−\mathrm{1}\right)\left(−\mathrm{2}\right)}{\mathrm{2}!}\left(\frac{\mathrm{3}{x}}{\mathrm{5}}\right)^{\mathrm{2}} +\frac{\left(−\mathrm{1}\right)\left(−\mathrm{2}\right)\left(−\mathrm{3}\right)}{\mathrm{3}!}\left(\frac{\mathrm{3}{x}}{\mathrm{5}}\right)^{\mathrm{3}} +…\:\right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{1}}{\mathrm{5}}−\frac{\mathrm{3}{x}}{\mathrm{25}}+\frac{\mathrm{9}{x}^{\mathrm{2}} }{\mathrm{125}}−\frac{\mathrm{27}{x}^{\mathrm{3}} }{\mathrm{625}}+… \\ $$$${because}\:{the}\:{binomial}\:{of}\left(\mathrm{1}+{X}\right)^{{n}} \\ $$$${valid}\:{for}\:\mid{X}\mid<\mathrm{1}\:{then}\:\mid\frac{\mathrm{3}{x}}{\mathrm{5}}\mid<\mathrm{1} \\ $$$${the}\:{set}\:{of}\:{the}\:{values}\:{of}\:{x}\:{is}\:\left\{{x}:\:−\frac{\mathrm{5}}{\mathrm{3}}<{x}<\frac{\mathrm{5}}{\mathrm{3}}\:\right\} \\ $$
Answered by mr W last updated on 01/Feb/21
(1/(3x+5)) =(1/5)×(1/(1−(−((3x)/5)))) =(1/5)[1+(−((3x)/5))+(−((3x)/5))^2 +(−((3x)/5))^3 +...] =(1/5)(1−((3x)/5)+((9x^2 )/(25))−((27x^3 )/(125))+...) =(1/5)−((3x)/(25))+((9x^2 )/(125))−((27x^3 )/(625))+... with ∣−((3x)/5)∣<1 ⇒−(5/3)<x<(5/3)
$$\frac{\mathrm{1}}{\mathrm{3}{x}+\mathrm{5}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{5}}×\frac{\mathrm{1}}{\mathrm{1}−\left(−\frac{\mathrm{3}{x}}{\mathrm{5}}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{5}}\left[\mathrm{1}+\left(−\frac{\mathrm{3}{x}}{\mathrm{5}}\right)+\left(−\frac{\mathrm{3}{x}}{\mathrm{5}}\right)^{\mathrm{2}} +\left(−\frac{\mathrm{3}{x}}{\mathrm{5}}\right)^{\mathrm{3}} +…\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{5}}\left(\mathrm{1}−\frac{\mathrm{3}{x}}{\mathrm{5}}+\frac{\mathrm{9}{x}^{\mathrm{2}} }{\mathrm{25}}−\frac{\mathrm{27}{x}^{\mathrm{3}} }{\mathrm{125}}+…\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{5}}−\frac{\mathrm{3}{x}}{\mathrm{25}}+\frac{\mathrm{9}{x}^{\mathrm{2}} }{\mathrm{125}}−\frac{\mathrm{27}{x}^{\mathrm{3}} }{\mathrm{625}}+… \\ $$$${with}\:\mid−\frac{\mathrm{3}{x}}{\mathrm{5}}\mid<\mathrm{1} \\ $$$$\Rightarrow−\frac{\mathrm{5}}{\mathrm{3}}<{x}<\frac{\mathrm{5}}{\mathrm{3}} \\ $$

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