# How do you find the sum of the finite geometric sequence of sum_(n=0)^50 10(2/3)^(n-1)?

May 18, 2018

$S = 30 \left[\frac{3}{2} + {\left(\frac{2}{3}\right)}^{50}\right]$

#### Explanation:

We know that,

"The sum of the first n-term of geometric sequence is :

$\textcolor{red}{{S}_{n} = \frac{a \left(1 - {r}^{n}\right)}{1 - r}} , w h e r e , a =$ first term, $r =$ common ratio "

We take,

$S = {\sum}_{n = 0}^{50} 10 {\left(\frac{2}{3}\right)}^{n - 1} = 10 {\sum}_{n = 0}^{50} {\left(\frac{2}{3}\right)}^{n - 1}$

$= 10 {\sum}_{n = 0} {\left(\frac{2}{3}\right)}^{n - 1} + 10 {\sum}_{n = 1}^{50} {\left(\frac{2}{3}\right)}^{n - 1}$

=$10 {\left(\frac{2}{3}\right)}^{-} 1 + 10 \left[{\left(\frac{2}{3}\right)}^{0} + {\left(\frac{2}{3}\right)}^{1} + {\left(\frac{2}{3}\right)}^{2} + \ldots + {\left(\frac{2}{3}\right)}^{49}\right]$

$= 10 \left(\frac{3}{2}\right) + 10 \left[1 + {\left(\frac{2}{3}\right)}^{1} + {\left(\frac{2}{3}\right)}^{2} + \ldots + {\left(\frac{2}{3}\right)}^{49}\right]$

$\therefore S = 10 \left(\frac{3}{2}\right) + 10 \cdot S ' \ldots \to \left(I\right)$

Where, $S ' = 1 + {\left(\frac{2}{3}\right)}^{1} + {\left(\frac{2}{3}\right)}^{2} + {\left(\frac{2}{3}\right)}^{3} + \ldots + {\left(\frac{2}{3}\right)}^{49}$

Here,

$a = 1 , r = \frac{2}{3} \mathmr{and} n = 1 + 49 = 50$

Using , color(red)(S_n=(a(1-r^n))/(1-r)

$\therefore S ' = {S}_{50} = \frac{1 \left(1 - {\left(\frac{2}{3}\right)}^{50}\right)}{1 - \frac{2}{3}}$

=>S'=(1-(2/3)^50)/((3-2)/3

$\implies S ' = \frac{1 - {\left(\frac{2}{3}\right)}^{50}}{\frac{1}{3}}$

$\implies S ' = 3 \left(1 - {\left(\frac{2}{3}\right)}^{50}\right)$

Thus, from $\left(I\right)$ we have

$S = 10 \left(\frac{3}{2}\right) + 10 \left[3 \left(1 - {\left(\frac{2}{3}\right)}^{50}\right)\right]$

$= \frac{30}{2} + 30 \left[1 - {\left(\frac{2}{3}\right)}^{50}\right]$

$= 30 \left[\frac{1}{2} + 1 - {\left(\frac{2}{3}\right)}^{50}\right]$

$= 30 \left[\frac{3}{2} + {\left(\frac{2}{3}\right)}^{50}\right]$