Chapter 12.3, Problem 40AYU

To determine

To find: The sum of series.

Answer to Problem 40AYU

The sum of series is $S_n=\frac{\left(3^n-1\right)}{6}$ .

Explanation of Solution

Given information:

The series is $\frac{3}{9}+\frac{3^2}{9}+\frac{3^3}{9}+\mathrm{............}+\frac{3^n}{9}$

Calculation:

For the given series $\begin{array}{c}{a}_1=\frac{3}{9}\\ =\frac{1}{3}\end{array}$ , $a_2=\frac{3^2}{9}$ and $a_3=\frac{3^3}{9}$ .

Calculate $\frac{a_2}{a_1}$ .

  $\begin{array}{c}\frac{a_2}{a_1}=\frac{\frac{3^2}{9}}{\frac{3}{9}}\\ =3\end{array}$

Calculate $\frac{a_3}{a_1}$ .

  $\begin{array}{c}\frac{a_3}{a_1}=\frac{\frac{3^3}{9}}{\frac{3^2}{9}}\\ =3\end{array}$

Since the ratio of successive terms is same it is a geometric series with common ratio $r=3$ .

The sum of first $n$ terms of a geometric sequence is calculated by,

  $S_n=a_1\frac{r^n-1}{r-1}$

Substitute the given values.

  $\begin{array}{l}{S}_n=\frac{\frac{1}{3}\left(3^n-1\right)}{3-1}\\ S_n=\frac{1}{3}\frac{\left(3^n-1\right)}{2}\\ S_n=\frac{\left(3^n-1\right)}{6}\end{array}$

Therefore, the sum of series is $S_n=\frac{\left(3^n-1\right)}{6}$ .