Chapter 12.1, Problem 35E

To determine

Calculate the first six partial sums $S_1,S_2,S_3,S_4,S_5,S_6$ of the given sequence $\frac{1}{3},\frac{1}{3^2},\frac{1}{3^3},\frac{1}{3^4},\mathrm{....}$ .

Answer to Problem 35E

The sums of the first six partial sequence is $S_1=\frac{1}{3},S_2=\frac{4}{9},S_3=\frac{13}{27},S_4=\frac{40}{81},S_5=\frac{121}{243},S_6=\frac{364}{729}$ .

Explanation of Solution

Given information:

It is given in the question that,

The sequence is $\frac{1}{3},\frac{1}{3^2},\frac{1}{3^3},\frac{1}{3^4},\mathrm{....}$ .

Formula used:

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

Calculation: The sequence is $\frac{1}{3},\frac{1}{3^2},\frac{1}{3^3},\frac{1}{3^4},\frac{1}{3^5},\frac{1}{3^6},\frac{1}{3^7},\mathrm{....}$ .

So,

  $\begin{array}{l}{S}_1=\frac{1}{3}\\ S_2=\frac{1}{3}+\frac{1}{3^2}=\frac{1}{3}+\frac{1}{9}=\frac{3+1}{9}=\frac{4}{9}\\ S_3=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}=\frac{9+3+1}{27}=\frac{13}{27}\\ S_4=1\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}=\frac{27+9+3+1}{81}=\frac{40}{81}\\ S_5=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}=\frac{81+27+9+3+1}{243}=\frac{121}{243}\\ S_6=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}+\frac{1}{3^6}=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}=\frac{243+81+27+9+3+1}{729}=\frac{364}{729}\end{array}$

Finally, $S_1=\frac{1}{3},S_2=\frac{4}{9},S_3=\frac{13}{27},S_4=\frac{40}{81},S_5=\frac{121}{243},S_6=\frac{364}{729}$ .