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Question-173124




Question Number 173124 by peter frank last updated on 06/Jul/22
Commented by aleks041103 last updated on 06/Jul/22
For which...what?
$${For}\:{which}…{what}? \\ $$
Commented by peter frank last updated on 07/Jul/22
Σ_(f=0) ^∞  (((x−3)/(x+1)))=(3/4)
$$\sum_{\mathrm{f}=\mathrm{0}} ^{\infty} \:\left(\frac{\mathrm{x}−\mathrm{3}}{\mathrm{x}+\mathrm{1}}\right)=\frac{\mathrm{3}}{\mathrm{4}} \\ $$
Answered by Mathspace last updated on 06/Jul/22
if ∣((x−3)/(x+1))∣<1  and x≠−1 ⇒  Σ_(n=0) ^∞ (((x−3)/(x+1)))^n =(1/(1−((x−3)/(x+1))))  =((x+1)/(x+1−x+3))=((x+1)/4)  if ∣((x−3)/(x+1))∣>1 the serie is div.  x=3 ⇒Σ(...)=0
$${if}\:\mid\frac{{x}−\mathrm{3}}{{x}+\mathrm{1}}\mid<\mathrm{1}\:\:{and}\:{x}\neq−\mathrm{1}\:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{0}} ^{\infty} \left(\frac{{x}−\mathrm{3}}{{x}+\mathrm{1}}\right)^{{n}} =\frac{\mathrm{1}}{\mathrm{1}−\frac{{x}−\mathrm{3}}{{x}+\mathrm{1}}} \\ $$$$=\frac{{x}+\mathrm{1}}{{x}+\mathrm{1}−{x}+\mathrm{3}}=\frac{{x}+\mathrm{1}}{\mathrm{4}} \\ $$$${if}\:\mid\frac{{x}−\mathrm{3}}{{x}+\mathrm{1}}\mid>\mathrm{1}\:{the}\:{serie}\:{is}\:{div}. \\ $$$${x}=\mathrm{3}\:\Rightarrow\Sigma\left(…\right)=\mathrm{0} \\ $$
Answered by aleks041103 last updated on 07/Jul/22
let a=((x−3)/(x+1))  ⇒Σ_(r=0) ^∞ a^r =(1/(1−a))=(3/4)  ⇒1−a=(4/3)  a=1−(4/3)=−(1/3)=a  since ∣a∣<1, the sum really converges  ((x−3)/(x+1))=((−1)/3)  ⇒3(x−3)=−(x+1)  3x−9=−x−1  4x−9+1=0  4x−8=0  x−2=0  x=2
$${let}\:{a}=\frac{{x}−\mathrm{3}}{{x}+\mathrm{1}} \\ $$$$\Rightarrow\underset{{r}=\mathrm{0}} {\overset{\infty} {\sum}}{a}^{{r}} =\frac{\mathrm{1}}{\mathrm{1}−{a}}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\Rightarrow\mathrm{1}−{a}=\frac{\mathrm{4}}{\mathrm{3}} \\ $$$${a}=\mathrm{1}−\frac{\mathrm{4}}{\mathrm{3}}=−\frac{\mathrm{1}}{\mathrm{3}}={a} \\ $$$${since}\:\mid{a}\mid<\mathrm{1},\:{the}\:{sum}\:{really}\:{converges} \\ $$$$\frac{{x}−\mathrm{3}}{{x}+\mathrm{1}}=\frac{−\mathrm{1}}{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{3}\left({x}−\mathrm{3}\right)=−\left({x}+\mathrm{1}\right) \\ $$$$\mathrm{3}{x}−\mathrm{9}=−{x}−\mathrm{1} \\ $$$$\mathrm{4}{x}−\mathrm{9}+\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{4}{x}−\mathrm{8}=\mathrm{0} \\ $$$${x}−\mathrm{2}=\mathrm{0} \\ $$$${x}=\mathrm{2} \\ $$

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