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Question Number 173124 by peter frank last updated on 06/Jul/22

Commented by aleks041103 last updated on 06/Jul/22

$$Forwhich...what?$$

Commented by peter frank last updated on 07/Jul/22

$$\sum _{\mathrm{f}=0}^{\mathrm{\infty }}\left(\frac{\mathrm{x}-3}{\mathrm{x}+1}\right)=\frac{3}{4}$$

Answered by Mathspace last updated on 06/Jul/22

$$if\mid \frac{x-3}{x+1}\mid <1andx\ne -1\Rightarrow$$$$\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{x-3}{x+1}\right)}^n=\frac{1}{1-\frac{x-3}{x+1}}$$$$=\frac{x+1}{x+1-x+3}=\frac{x+1}{4}$$$$if\mid \frac{x-3}{x+1}\mid >1theserieisdiv.$$$$x=3\Rightarrow \mathrm{\Sigma }(...)=0$$

Answered by aleks041103 last updated on 07/Jul/22

$$leta=\frac{x-3}{x+1}$$$$\Rightarrow \underset{r=0}{\sum ^{\mathrm{\infty }}}{a}^r=\frac{1}{1-a}=\frac{3}{4}$$$$\Rightarrow 1-a=\frac{4}{3}$$$$a=1-\frac{4}{3}=-\frac{1}{3}=a$$$$since\mid a\mid <1,thesumreallyconverges$$$$\frac{x-3}{x+1}=\frac{-1}{3}$$$$\Rightarrow 3(x-3)=-(x+1)$$$$3x-9=-x-1$$$$4x-9+1=0$$$$4x-8=0$$$$x-2=0$$$$x=2$$

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