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let-S-x-n-0-3x-n-2-using-the-sum-above-find-n-0-1-n-1-3-n-1-n-3-

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Question Number 159379 by HongKing last updated on 16/Nov/21
let S(x) =Σ_(n=0) ^∞ (3x)^(n+2) using the sum above find: Σ_(n=0) ^∞ (((-1)^(n+1) )/(3^(n+1) (n + 3)))
letS(x)=∑∞n=0(3x)n+2usingthesumabovefind:∑∞n=0(−1)n+13n+1(n+3)
Answered by Ar Brandon last updated on 16/Nov/21
Σ_(n=0) ^∞ (−(1/3))^(n+1) (1/(n+3))=Σ_(n=0) ^∞ (−(1/3))^(n+1) ∫_0 ^1 x^(n+2) dx =−3Σ_(n=0) ^∞ (−(1/3))^(n+2) ∫_0 ^1 x^(n+2) dx=−3∫_0 ^1 Σ_(n=0) ^∞ (−(x/3))^(n+2) dx =−3∫_0 ^1 ((x^2 /9)/(1+(x/3)))dx=−∫_0 ^1 (x^2 /(3+x))dx=−∫_0 ^1 (((x+3)^2 −6(x+3)+9)/(x+3))dx =−[(x^2 /2)+3x−6x+9ln(x+3)]_0 ^1 =9ln3+(5/2)−9ln4=(5/2)+9ln((3/4))
∑∞n=0(−13)n+11n+3=∑∞n=0(−13)n+1∫01xn+2dx=−3∑∞n=0(−13)n+2∫01xn+2dx=−3∫01∑∞n=0(−x3)n+2dx=−3∫01x291+x3dx=−∫01x23+xdx=−∫01(x+3)2−6(x+3)+9x+3dx=−[x22+3x−6x+9ln(x+3)]01=9ln3+52−9ln4=52+9ln(34)
Commented by HongKing last updated on 16/Nov/21
thank you so much my dear Ser cool
thankyousomuchmydearSercool
Commented by HongKing last updated on 16/Nov/21
my dear Ser, but they said use the first sum S(x) to evaluate the new one
mydearSer,buttheysaidusethefirstsumS(x)toevaluatethenewone
Commented by Ar Brandon last updated on 16/Nov/21
Answered by Ar Brandon last updated on 16/Nov/21
S(x)=Σ_(n=0) ^∞ (3x)^(n+2) =((9x^2 )/(1−3x))=(((1−3x)^2 −2(1−3x)+1)/(1−3x)) ∫S(x)dx=(1/3)Σ_(n=0) ^∞ (((3x)^(n+3) )/(n+3))+C=x−(3/2)x^2 −2x−((ln(1−3x))/3)+C ∫S(0)dx=0=C⇒∫S(x)dx=(1/3)Σ_(n=0) ^∞ (((3x)^(n+3) )/(n+3))=−x−(3/2)x^2 −((ln(1−3x))/3) ∫S(−(1/9))dx=(1/3)Σ_(n=0) ^∞ (((−1)^(n+3) )/(3^(n+3) (n+3)))=(1/9)−(1/(54))−(1/3)ln((4/3)) ⇒Σ_(n=0) ^∞ (((−1)^(n+1) )/(3^(n+1) (n+3)))=3−(1/2)−9ln((4/3))=(5/2)−9ln((4/3))
S(x)=∑∞n=0(3x)n+2=9x21−3x=(1−3x)2−2(1−3x)+11−3x∫S(x)dx=13∑∞n=0(3x)n+3n+3+C=x−32x2−2x−ln(1−3x)3+C∫S(0)dx=0=C⇒∫S(x)dx=13∑∞n=0(3x)n+3n+3=−x−32x2−ln(1−3x)3∫S(−19)dx=13∑∞n=0(−1)n+33n+3(n+3)=19−154−13ln(43)⇒∑∞n=0(−1)n+13n+1(n+3)=3−12−9ln(43)=52−9ln(43)
Commented by HongKing last updated on 16/Nov/21
perfect, thank you so much my dear Ser
perfect,thankyousomuchmydearSer
Commented by Ar Brandon last updated on 16/Nov/21
You're welcome, Sir.

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