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

Jun 10, 2016

#### Explanation:

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

${n}^{t h}$ of such a series is $a {r}^{n - 1}$ and sum of the series ${S}_{n}$ up to $n$ terms is given by

${S}_{n} = a \frac{{r}^{n} - 1}{r - 1}$ (used if $| r | > 1$)

or ${S}_{n} = a \frac{1 - {r}^{n}}{1 - r}$ (used if $| r | < 1$)

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

Here $a = \frac{1}{3}$ and $r = \frac{\frac{1}{9}}{\frac{1}{3}} = \frac{1}{9} \times \frac{3}{1} = \frac{1}{3}$

Hence, ${n}^{t h}$ term of the series is $\frac{1}{3} {\left(\frac{1}{3}\right)}^{n - 1} = \frac{1}{3} ^ n$ and sum of the series ${S}_{n}$ is

${S}_{n} = \frac{1}{3} \frac{\left(1 - {\left(\frac{1}{3}\right)}^{n}\right)}{\left(1 - \frac{1}{3}\right)} = \frac{1}{3} \frac{\left(1 - \left(\frac{1}{3} ^ n\right)\right)}{\frac{2}{3}}$

= $\frac{1}{3} \left(1 - \left(\frac{1}{3} ^ n\right)\right) \times \frac{3}{2}$

= 1/2(1-(1/3^n)

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