Chapter 11.6, Problem 29WE

a.

To determine

To calculate: The sum of series

Answer to Problem 29WE

The sum of the infinity series is 3.

Explanation of Solution

Given information: The series is $1+\frac{2}{3}+\frac{4}{9}+\frac{8}{27}+\mathrm{....}$

Hence, the given series is in G.P

The first term, $a=1$

Second terms, $a_1=\frac{2}{3}$

The common ratio, $r=\frac{a_1}{a}=\frac{\frac{2}{3}}{1}=\frac{2}{3}$

As we know sum of infinity term of a G.P is $S=\frac{a}{1-r}$ .

Calculation:

  $\begin{array}{l}=\frac{1}{1-\frac{2}{3}}\\ =\frac{1}{\frac{3-2}{3}}\\ =3\end{array}$

Hence, the sum of the infinity series is 3.

b.

To determine

To calculate:

Answer to Problem 29WE

The numbers of terms are 31.

Explanation of Solution

Given information: The series is $1+\frac{2}{3}+\frac{4}{9}+\frac{8}{27}+\mathrm{....}$

Hence, the given series is in G.P

The first term, $a=1$

Second terms, $a_1=\frac{2}{3}$

The common ratio, $r=\frac{a_1}{a}=\frac{\frac{2}{3}}{1}=\frac{2}{3}$

Formula used: Sum of G.P $S_n=\frac{a\left(1-r^n\right)}{1-r}\because \left[r<1\right]$ , where n is numbers of terms

Calculation:

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

Now,

Now, numbers of terms

  $\therefore S_n=S$

  $\begin{array}{l}\therefore 3\left\{1-{\left(\frac{2}{3}\right)}^n\right\}=3\\ \Rightarrow 1-{\left(\frac{2}{3}\right)}^n=1\\ \Rightarrow S-S_n=3-3\left\{1-{\left(\frac{2}{3}\right)}^n\right\}\\ \Rightarrow S-S_n=3\left\{1-1-{\left(\frac{2}{3}\right)}^n\right\}\\ \Rightarrow S-S_n={\left(\frac{2}{3}\right)}^n=-\frac{1}{300000}\\ \therefore {\left(\frac{2}{3}\right)}^n=\frac{1}{300000}:n=\frac{\ln \left(300000\right)}{\ln \left(\frac{2}{3}\right)}\left(\text{Decimal}:n=31.10387\dots \right)\end{array}$

Hence, the numbers of terms are 31.