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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 .
Findthefirstfourtermsintheseriesexpansionof13x+5ascendingpowerxandstatethesetofvaluesofxforwhichthisexpansionisvalid.
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) }
remarkbinomialtheoremforn∉Nand∣X∣<1is(1+X)n=1+nX+n(n−1)2!X2+n(n−1)(n−2)3!X3+…thentransform13x+5intoform(1+X)nbywritting13x+5=15(1+3x/5)=15(1+3x5)−1usingn=−1,X=3x513x+5=15{1+(−1)3x5+(−1)(−2)2!(3x5)2+(−1)(−2)(−3)3!(3x5)3+…}=15−3x25+9x2125−27x3625+…becausethebinomialof(1+X)nvalidfor∣X∣<1then∣3x5∣<1thesetofthevaluesofxis{x:−53<x<53}
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)
13x+5=15×11−(−3x5)=15[1+(−3x5)+(−3x5)2+(−3x5)3+…]=15(1−3x5+9x225−27x3125+…)=15−3x25+9x2125−27x3625+…with∣−3x5∣<1⇒−53<x<53

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