Chapter 12, Problem 9T

(a)

To determine

To find: The sum of the given geometric sequence.

Answer to Problem 9T

The sum of the given geometric sequence is $\frac{58025}{59049}$.

Explanation of Solution

Given:

The given sequence is $\frac{1}{3}+\frac{2}{3^2}+\frac{2^2}{3^3}+\frac{2^3}{3^4}+\cdots +\frac{2^9}{3^{10}}$

Formula used:

For an infinite geometric series in the form, $a+ar+a{r}^2+a{r}^3+a{r}^4+\cdots +a{r}^{n-1}+a{r}^n$, the $n\text{th}$ partial sum of such a series is given by the formulae,

$S_n=a\left(\frac{1-r^n}{1-r}\right)$

Where, a is the first term and r is the common ratio of the sequence.

Calculation:

For an infinite geometric series in the form, $a+ar+a{r}^2+a{r}^3+a{r}^4+\cdots +a{r}^{n-1}+a{r}^n$, the $n\text{th}$ partial sum of such a series is given by the formulae,

$S_n=a\left(\frac{1-r^n}{1-r}\right)$ (1)

Compare the given sequence is $\frac{1}{3}+\frac{2}{3^2}+\frac{2^2}{3^3}+\frac{2^3}{3^4}+\cdots +\frac{2^9}{3^{10}}$ with the general geometric series in the form $a+ar+a{r}^2+a{r}^3+a{r}^4+\cdots +a{r}^{n-1}+a{r}^n$.

The value of the term $a$ is $\frac{1}{3}$.

For geometric ratio $\left(r\right)$ of given geometric sequence the ratio of succeeding term to that of preceding term is defined as geometric ratio $\left(r\right)$.

$\begin{array}{c}r=\left(\frac{2/3^2}{1/3}\right)\\ =\frac{2}{3}\end{array}$

The value of geometric ratio $\left(r\right)$ for the given sequence is $\frac{2}{3}$.

As, in the geometric sequence $\frac{1}{3}+\frac{2}{3^2}+\frac{2^2}{3^3}+\frac{2^3}{3^4}+\cdots +\frac{2^9}{3^{10}}$ there are 10 terms, therefore the value of n is 10.

Substitute $\frac{2}{3}$ for $\left(r\right)$, $\frac{1}{3}$ for $a$ and 10 for n in equation (1) to sum of the given geometric sequence.

$\begin{array}{l}{S}_{10}=\frac{1}{3}\left(\frac{1-{\left(\frac{2}{3}\right)}^{10}}{1-\left(\frac{2}{3}\right)}\right)\\ =\frac{1}{3}\left(\frac{1-\frac{1024}{59049}}{1-\left(\frac{2}{3}\right)}\right)\\ =\frac{1}{3}\left(\frac{\frac{59049-1024}{59049}}{\left(\frac{3-2}{3}\right)}\right)\\ =\frac{1}{3}\left(\frac{58025\times 3}{59049}\right)\end{array}$

Further simplify of above equation,

$\begin{array}{c}{S}_{10}=\frac{1}{3}\left(\frac{58025\times 3}{59049}\right)\\ =\frac{58025}{59049}\end{array}$

Thus, the sum of the given geometric sequence is $\frac{58025}{59049}$.

(b)

To determine

To find: The sum of the given geometric sequence.

Answer to Problem 9T

The sum of the given geometric sequence is $2+\sqrt{2}$.

Explanation of Solution

Given:

The given sequence is $1+\frac{1}{2^{1/2}}+\frac{1}{2}+\frac{1}{2^{3/2}}+\cdots$

Definition used:

For an infinite geometric series in the form, $a+ar+a{r}^2+a{r}^3+a{r}^4+\cdots +a{r}^{n-1}+a{r}^n$, the $n\text{th}$ partial sum of such a series is given by the formulae,

$S_n=\frac{a}{1-r}$

Where, a is the first term and r is the common ratio of the sequence.

Calculation:

For an infinite geometric series in the form, $a+ar+a{r}^2+a{r}^3+a{r}^4+\cdots +a{r}^{n-1}+a{r}^n$, the $n\text{th}$ partial sum of such a series is given by the formulae,

$S_n=\frac{a}{1-r}$ (1)

Compare the given sequence is $1+\frac{1}{2^{1/2}}+\frac{1}{2}+\frac{1}{2^{3/2}}+\cdots$ with the general geometric series in the form $a+ar+a{r}^2+a{r}^3+a{r}^4+\cdots +a{r}^{n-1}+a{r}^n$.

The value of the term $a$ is $1$.

For geometric ratio $\left(r\right)$ of given geometric sequence the ratio of succeeding term to that of preceding term is defined as geometric ratio $\left(r\right)$.

$\begin{array}{c}r=\left(\frac{\frac{1}{2^{1/2}}}{1}\right)\\ =\frac{1}{2^{1/2}}\end{array}$

The value of geometric ratio $\left(r\right)$ for the given sequence is $\frac{1}{2^{1/2}}$.

As, in the geometric sequence $1+\frac{1}{2^{1/2}}+\frac{1}{2}+\frac{1}{2^{3/2}}+\cdots$ there are n number of terms.

Substitute $\frac{1}{2^{1/2}}$ for $\left(r\right)$, and $1$ for $a$ in equation (1) to sum of the given geometric sequence.

$\begin{array}{c}{S}_n=\frac{1}{1-\frac{1}{2^{1/2}}}\\ =\frac{1}{\frac{2^{1/2}-1}{2^{1/2}}}\\ =\frac{2^{1/2}}{2^{1/2}-1}\end{array}$

Multiply and divide the above equation with $2^{1/2}+1$ for simplified term,

$\begin{array}{c}{S}_n=\frac{2^{1/2}}{2^{1/2}-1}\times \frac{2^{1/2}+1}{2^{1/2}+1}\\ =\frac{2^{1/2}\left(2^{1/2}+1\right)}{{\left(2^{1/2}\right)}^2-1^2}\\ =\frac{2+2^{1/2}}{2-1}\\ =2+\sqrt{2}\end{array}$

Thus, the sum of the given geometric sequence is $2+\sqrt{2}$.