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Question Number 195129 by a.lgnaoui last updated on 25/Jul/23

Calculer la valeur de la serie suivante: S=(3/2)+(5/8)+(7/(32))+(9/(128))+.....

$$\mathrm{Calculer}\:\mathrm{la}\:\mathrm{valeur}\:\mathrm{de}\:\mathrm{la}\:\mathrm{serie}\:\mathrm{suivante}: \\ $$$$\boldsymbol{\mathrm{S}}=\frac{\mathrm{3}}{\mathrm{2}}+\frac{\mathrm{5}}{\mathrm{8}}+\frac{\mathrm{7}}{\mathrm{32}}+\frac{\mathrm{9}}{\mathrm{128}}+..... \\ $$

Answered by MM42 last updated on 25/Jul/23

(1/2^2 )S=(3/2^3 )+(5/2^5 )+(7/2^7 )+... ⇒(1−(1/4))S=(3/2)+2((1/2^3 )+(1/2^5 )+...) =(3/2)+2(((1/8)/(1−(1/4))))=(3/2)+(1/3)=((11)/6) ✓

$$\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }{S}=\frac{\mathrm{3}}{\mathrm{2}^{\mathrm{3}} }+\frac{\mathrm{5}}{\mathrm{2}^{\mathrm{5}} }+\frac{\mathrm{7}}{\mathrm{2}^{\mathrm{7}} }+... \\ $$$$\Rightarrow\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}\right){S}=\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{5}} }+...\right) \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{2}\left(\frac{\frac{\mathrm{1}}{\mathrm{8}}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}}\right)=\frac{\mathrm{3}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{11}}{\mathrm{6}}\:\checkmark \\ $$

Commented by a.lgnaoui last updated on 25/Jul/23

thanks

$$\mathrm{thanks} \\ $$

Commented by a.lgnaoui last updated on 25/Jul/23

but the answer is wrong

$$\mathrm{but}\:\:\mathrm{the}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{wrong} \\ $$

Commented by MM42 last updated on 25/Jul/23

yes (3/4)S=((11)/6)⇒S=((11)/6)×(4/3)=((22)/9)

$${yes} \\ $$$$\frac{\mathrm{3}}{\mathrm{4}}{S}=\frac{\mathrm{11}}{\mathrm{6}}\Rightarrow{S}=\frac{\mathrm{11}}{\mathrm{6}}×\frac{\mathrm{4}}{\mathrm{3}}=\frac{\mathrm{22}}{\mathrm{9}} \\ $$

Answered by BaliramKumar last updated on 25/Jul/23

S=(3/2)+(5/8)+(7/(32))+(9/(128))+..... (i) (S/4) = (3/8)+(5/(32))+(7/(128))+........... (ii) (i) − (ii) S−(S/4) = (3/2)+((5/8)−(3/8))+((7/(32))−(5/(32)))+((9/(128))−(7/(128)))+....... ((3S)/4) = (3/2)+((1/4))+((1/(16)))+((1/(64)))+....... ((3S)/4) = (3/2)+ ((1/4)/(1−(1/4))) ((3S)/4) = (3/2)+ (1/3) = ((11)/6) S = ((11×4)/(6×3)) S =((22)/9)

$$\boldsymbol{\mathrm{S}}=\frac{\mathrm{3}}{\mathrm{2}}+\frac{\mathrm{5}}{\mathrm{8}}+\frac{\mathrm{7}}{\mathrm{32}}+\frac{\mathrm{9}}{\mathrm{128}}+.....\:\:\:\:\:\:\:\:\:\left(\mathrm{i}\right) \\ $$$$\frac{\mathrm{S}}{\mathrm{4}}\:=\:\frac{\mathrm{3}}{\mathrm{8}}+\frac{\mathrm{5}}{\mathrm{32}}+\frac{\mathrm{7}}{\mathrm{128}}+...........\:\:\:\:\:\:\:\:\:\:\left(\mathrm{ii}\right) \\ $$$$\left(\mathrm{i}\right)\:−\:\left(\mathrm{ii}\right) \\ $$$$\mathrm{S}−\frac{\mathrm{S}}{\mathrm{4}}\:=\:\frac{\mathrm{3}}{\mathrm{2}}+\left(\frac{\mathrm{5}}{\mathrm{8}}−\frac{\mathrm{3}}{\mathrm{8}}\right)+\left(\frac{\mathrm{7}}{\mathrm{32}}−\frac{\mathrm{5}}{\mathrm{32}}\right)+\left(\frac{\mathrm{9}}{\mathrm{128}}−\frac{\mathrm{7}}{\mathrm{128}}\right)+....... \\ $$$$\frac{\mathrm{3S}}{\mathrm{4}}\:=\:\frac{\mathrm{3}}{\mathrm{2}}+\left(\frac{\mathrm{1}}{\mathrm{4}}\right)+\left(\frac{\mathrm{1}}{\mathrm{16}}\right)+\left(\frac{\mathrm{1}}{\mathrm{64}}\right)+....... \\ $$$$\frac{\mathrm{3S}}{\mathrm{4}}\:=\:\frac{\mathrm{3}}{\mathrm{2}}+\:\:\frac{\frac{\mathrm{1}}{\mathrm{4}}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}} \\ $$$$\frac{\mathrm{3S}}{\mathrm{4}}\:=\:\frac{\mathrm{3}}{\mathrm{2}}+\:\:\frac{\mathrm{1}}{\mathrm{3}}\:=\:\frac{\mathrm{11}}{\mathrm{6}} \\ $$$$\mathrm{S}\:=\:\frac{\mathrm{11}×\mathrm{4}}{\mathrm{6}×\mathrm{3}} \\ $$$$\mathrm{S}\:=\frac{\mathrm{22}}{\mathrm{9}} \\ $$$$ \\ $$$$ \\ $$

Commented by a.lgnaoui last updated on 25/Jul/23

Exact ,thanks

$$\mathrm{Exact}\:,\mathrm{thanks} \\ $$

Answered by a.lgnaoui last updated on 25/Jul/23

4S=6+(5/2)+(7/8)+(9/(32))+... 4S−S=6+1+(1/4)+(1/(16))+... 3S=6+(1+(1/4)+(1/4^2 )+....) =6+(1/(1−(1/4)))=6+(4/3)=((22)/3) ⇒ S=((22)/9)

$$\mathrm{4S}=\mathrm{6}+\frac{\mathrm{5}}{\mathrm{2}}+\frac{\mathrm{7}}{\mathrm{8}}+\frac{\mathrm{9}}{\mathrm{32}}+... \\ $$$$\mathrm{4S}−\mathrm{S}=\mathrm{6}+\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{16}}+... \\ $$$$\mathrm{3S}=\mathrm{6}+\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{2}} }+....\right) \\ $$$$\:\:\:=\mathrm{6}+\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}}=\mathrm{6}+\frac{\mathrm{4}}{\mathrm{3}}=\frac{\mathrm{22}}{\mathrm{3}} \\ $$$$\:\:\:\:\Rightarrow\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{S}}=\frac{\mathrm{22}}{\mathrm{9}} \\ $$

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