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Question Number 10671 by Saham last updated on 22/Feb/17

Σ_(n = 1) ^∞ 5((1/4))^(n − 1)

$$\underset{\mathrm{n}\:=\:\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{5}\left(\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{n}\:−\:\mathrm{1}} \\ $$

Answered by FilupS last updated on 22/Feb/17

S=5Σ_(n=1) ^∞ (1/4^(n−1) ) S=5((1/4^0 )+(1/4^1 )+(1/4^2 )+...) (1/5)S=((1/4^0 )+(1/4^1 )+(1/4^2 )+...) (1/5)S=(1/4^0 )+((1/4^1 )+(1/4^2 )+(1/4^3 )+...) (1/5)S=1+(1/4)((1/4^0 )+(1/4^1 )+(1/4^2 )+...) (1/5)S=1+(1/4)((1/5)S) (1/5)S=1+(1/(20))S ((1/5)−(1/(20)))S=1 (((20−5)/(100)))S=1 ((15)/(100))S=1 (3/(20))S=1 S=((20)/3)

$${S}=\mathrm{5}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{4}^{{n}−\mathrm{1}} } \\ $$$${S}=\mathrm{5}\left(\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{0}} }+\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{1}} }+\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{2}} }+...\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{5}}{S}=\left(\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{0}} }+\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{1}} }+\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{2}} }+...\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{5}}{S}=\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{0}} }+\left(\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{1}} }+\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{3}} }+...\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{5}}{S}=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{0}} }+\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{1}} }+\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{2}} }+...\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{5}}{S}=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{1}}{\mathrm{5}}{S}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{5}}{S}=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{20}}{S} \\ $$$$\left(\frac{\mathrm{1}}{\mathrm{5}}−\frac{\mathrm{1}}{\mathrm{20}}\right){S}=\mathrm{1} \\ $$$$\left(\frac{\mathrm{20}−\mathrm{5}}{\mathrm{100}}\right){S}=\mathrm{1} \\ $$$$\frac{\mathrm{15}}{\mathrm{100}}{S}=\mathrm{1} \\ $$$$\frac{\mathrm{3}}{\mathrm{20}}{S}=\mathrm{1} \\ $$$${S}=\frac{\mathrm{20}}{\mathrm{3}} \\ $$

Commented by Saham last updated on 22/Feb/17

wow, i really appreciate.

$$\mathrm{wow},\:\mathrm{i}\:\mathrm{really}\:\mathrm{appreciate}. \\ $$

Commented by Saham last updated on 22/Feb/17

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Answered by mrW1 last updated on 22/Feb/17

an other way: G.P. with a_1 =5 q=(1/4) Sn=((a_1 (1−q^n ))/(1−q))=((5(1−((1/4))^n ))/(1−(1/4)))=((20)/3)(1−(1/4^n )) lim_(n→0) S_n =((20)/3)

$${an}\:{other}\:{way}: \\ $$$$ \\ $$$${G}.{P}.\:{with} \\ $$$${a}_{\mathrm{1}} =\mathrm{5} \\ $$$${q}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${Sn}=\frac{{a}_{\mathrm{1}} \left(\mathrm{1}−{q}^{{n}} \right)}{\mathrm{1}−{q}}=\frac{\mathrm{5}\left(\mathrm{1}−\left(\frac{\mathrm{1}}{\mathrm{4}}\right)^{{n}} \right)}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}}=\frac{\mathrm{20}}{\mathrm{3}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}^{{n}} }\right) \\ $$$$\underset{{n}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{S}_{{n}} =\frac{\mathrm{20}}{\mathrm{3}} \\ $$

Commented by Saham last updated on 22/Feb/17

Great, God bless you sir.

$$\mathrm{Great},\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Commented by Saham last updated on 22/Feb/17

Thanks for your help.

$$\mathrm{Thanks}\:\mathrm{for}\:\mathrm{your}\:\mathrm{help}. \\ $$

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