Chapter 12.1, Problem 37E

To determine

Calculate the first four partial sums and also find the $n^{th}$ partial sum of the sequence $a_n$ .

Answer to Problem 37E

The first four partial sums is,

  $S_1=\frac{2}{3},S_2=\frac{8}{9},S_3=\frac{26}{27},S_4=\frac{80}{81}$

And, the $n^{th}$ partial sum of the sequence is,

  $S_n=\frac{3^n-1}{3^n}$

Explanation of Solution

Given information:

it is given in the question that, $a_n=\frac{2}{3^n}$ .

Formula used:

For finding the partial sums use the formula,

  $S_n=a_1+a_2+a_3+\mathrm{........}+a_n$ .

Calculation: Since, $a_n=\frac{2}{3^n}$ so now by putting the value of $n=1,2,3,4$ the first four term of $A.P$ are arrived

That is,

  $a_1=\frac{2}{3^1}=\frac{2}{3}$

  $a_2=\frac{2}{3^2}=\frac{2}{9}$

  $a_3=\frac{2}{3^3}=\frac{2}{27}$

  $a_4=\frac{2}{3^4}=\frac{2}{81}$

For finding the $n^{th}$ partial sum of the sequence use the formula $S_n=a_1+a_2+a_3+\mathrm{........}+a_n$

  $\begin{array}{l}{S}_1=\frac{2}{3}\\ S_2=\frac{2}{3}+\frac{2}{9}=\frac{6+2}{9}=\frac{8}{9}\\ S_3=\frac{2}{3}+\frac{2}{9}+\frac{2}{27}=\frac{18+6+2}{27}=\frac{26}{27}\\ S_4=\frac{2}{3}+\frac{2}{9}+\frac{2}{27}+\frac{2}{81}=\frac{54+18+6+2}{81}=\frac{80}{81}\end{array}$

By observing the above terms numerator is always $1$ less than the Denominator and denominator is always in the multiple of $3$ so the,

  $S_n=\frac{3^n-1}{3^n}$