n-1-5-1-4-n-1-

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

\underset{n = 1}{\sum^{\infty}} 5 {(\frac{1}{4})}^{n - 1}

Answered by FilupS last updated on 22/Feb/17

S = 5 \underset{n = 1}{\sum^{\infty}} \frac{1}{4^{n - 1}}

S = 5 (\frac{1}{4^{0}} + \frac{1}{4^{1}} + \frac{1}{4^{2}} + \dots)

\frac{1}{5} S = (\frac{1}{4^{0}} + \frac{1}{4^{1}} + \frac{1}{4^{2}} + \dots)

\frac{1}{5} S = \frac{1}{4^{0}} + (\frac{1}{4^{1}} + \frac{1}{4^{2}} + \frac{1}{4^{3}} + \dots)

\frac{1}{5} S = 1 + \frac{1}{4} (\frac{1}{4^{0}} + \frac{1}{4^{1}} + \frac{1}{4^{2}} + \dots)

\frac{1}{5} S = 1 + \frac{1}{4} (\frac{1}{5} S)

\frac{1}{5} S = 1 + \frac{1}{20} S

(\frac{1}{5} - \frac{1}{20}) S = 1

(\frac{20 - 5}{100}) S = 1

\frac{15}{100} S = 1

\frac{3}{20} S = 1

S = \frac{20}{3}

Commented by Saham last updated on 22/Feb/17

wow, i really appreciate .

Commented by Saham last updated on 22/Feb/17

God bless you sir .

Answered by mrW1 last updated on 22/Feb/17

a n o t h e r w a y :

G . P . w i t h

a_{1} = 5

q = \frac{1}{4}

S n = \frac{a_{1} (1 - q^{n})}{1 - q} = \frac{5 (1 - {(\frac{1}{4})}^{n})}{1 - \frac{1}{4}} = \frac{20}{3} (1 - \frac{1}{4^{n}})

\lim_{n \to 0} S_{n} = \frac{20}{3}

Commented by Saham last updated on 22/Feb/17

Great, God bless you sir .

Commented by Saham last updated on 22/Feb/17

Thanks for your help .

Facebook Tweet Pin