How do you find the Geometric formula for this sequence 1/3,1/9,1/27,1/81?

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How do you find the Geometric formula for this sequence 1/3,1/9,1/27,1/81?

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Answer 1 · 2016-06-10T09:16:16

Please see below.

Explanation:

A geometric series is written as ${a, a r, a r^{2}, a r^{3}, a r^{4}, .}$ ${a,ar,ar^2,ar^3,ar^4,.}$ , where $a$ $a$ is first term (of the series) and $r$ $r$ is the ratio of a term to its preceding term.

$n^{t h}$ $n^(th)$ of such a series is $a r^{n - 1}$ $ar^(n-1)$ and sum of the series $S_{n}$ $S_n$ up to $n$ $n$ terms is given by

$S_{n} = a \frac{r^{n} - 1}{r - 1}$ $S_n=a(r^n-1)/(r-1)$ (used if $| r | > 1$ $|r|>1$ )

or $S_{n} = a \frac{1 - r^{n}}{1 - r}$ $S_n=a(1-r^n)/(1-r)$ (used if $| r | < 1$ $|r|<1$ )

If $| r | < 1$ $|r|<1$ , sum of infinite series is $\frac{a}{1 - r}$ $a/(1-r)$

Here $a = \frac{1}{3}$ $a=1/3$ and $r = \frac{\frac{1}{9}}{\frac{1}{3}} = \frac{1}{9} \times \frac{3}{1} = \frac{1}{3}$ $r=(1/9)/(1/3)=1/9xx3/1=1/3$

Hence, $n^{t h}$ $n^(th)$ term of the series is $\frac{1}{3} {(\frac{1}{3})}^{n - 1} = \frac{1}{3^{n}}$ $1/3(1/3)^(n-1)=1/3^n$ and sum of the series $S_{n}$ $S_n$ is

$S_{n} = \frac{1}{3} \frac{(1 - {(\frac{1}{3})}^{n})}{(1 - \frac{1}{3})} = \frac{1}{3} \frac{(1 - (\frac{1}{3^{n}}))}{\frac{2}{3}}$ $S_n=1/3((1-(1/3)^n))/((1-1/3))=1/3((1-(1/3^n)))/(2/3)$

= $\frac{1}{3} (1 - (\frac{1}{3^{n}})) \times \frac{3}{2}$ $1/3(1-(1/3^n))xx3/2$

= $\frac{1}{2} (1 - (\frac{1}{3^{n}})$ $1/2(1-(1/3^n)$

And sum of infinite series is $\frac{1}{2}$ $1/2$

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How do you find the Geometric formula for this sequence 1/3,1/9,1/27,1/81?

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