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Question Number 92324 by I want to learn more last updated on 06/May/20

Find the value of  x  for which    Σ_(n  =  0) ^(n  =  ∞)  16((3/4)x  +  1)^n   (a)   Is convergent  (b)   Is equal to  10(2/3)

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\:\mathrm{x}\:\:\mathrm{for}\:\mathrm{which}\:\:\:\:\underset{\mathrm{n}\:\:=\:\:\mathrm{0}} {\overset{\mathrm{n}\:\:=\:\:\infty} {\sum}}\:\mathrm{16}\left(\frac{\mathrm{3}}{\mathrm{4}}\mathrm{x}\:\:+\:\:\mathrm{1}\right)^{\mathrm{n}} \\ $$$$\left(\mathrm{a}\right)\:\:\:\mathrm{Is}\:\mathrm{convergent} \\ $$$$\left(\mathrm{b}\right)\:\:\:\mathrm{Is}\:\mathrm{equal}\:\mathrm{to}\:\:\mathrm{10}\frac{\mathrm{2}}{\mathrm{3}} \\ $$

Answered by mr W last updated on 06/May/20

convergent if  ∣(3/4)x+1∣<1  −1<(3/4)x+1<1  ⇒−(8/3)<x<0    S=16×(1/(1−((3/4)x+1)))=((32)/3)  ⇒x=−2

$${convergent}\:{if} \\ $$$$\mid\frac{\mathrm{3}}{\mathrm{4}}{x}+\mathrm{1}\mid<\mathrm{1} \\ $$$$−\mathrm{1}<\frac{\mathrm{3}}{\mathrm{4}}{x}+\mathrm{1}<\mathrm{1} \\ $$$$\Rightarrow−\frac{\mathrm{8}}{\mathrm{3}}<{x}<\mathrm{0} \\ $$$$ \\ $$$${S}=\mathrm{16}×\frac{\mathrm{1}}{\mathrm{1}−\left(\frac{\mathrm{3}}{\mathrm{4}}{x}+\mathrm{1}\right)}=\frac{\mathrm{32}}{\mathrm{3}} \\ $$$$\Rightarrow{x}=−\mathrm{2} \\ $$

Commented by I want to learn more last updated on 06/May/20

Thanks sir

$$\mathrm{Thanks}\:\mathrm{sir} \\ $$

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