Chapter 7.3, Problem 13Q

Solution

To calculate : The rule for $n^{th}$ term $a_n$ of the arithmetic or geometric series, value of $a_{15}$ and the sum of first 15 terms of the sequence $2,8,32,128,\dots$ .

The rule for $n^{th}$ term of the sequence $2,8,32,128,\dots$ is $a_n=2^{2n-1}$ , value of $a_{15}$ is $536870912$ and the sum of first 15 terms is $715827882$

Given information :

The sequence $2,8,32,128,\dots$ .

Formula used :

The $n^{th}$ term of an arithmetic series with first term as $a_1$ and common difference $d$ is given by $a_n=a_1+\left(n-1\right)d$ .

The sum of first $n$ terms of an arithmetic series is $S_n=n\left(\frac{a_1+a_n}{2}\right)$ .

The $n^{th}$ term of a geometric series with first term as $a_1$ and common ratio $r$ is given by $a_n=a_1{r}^{n-1}$ .

The sum of the first $n$ terms of geometric series with first term $a_1$ and common ration $r$ is calculated using the formula $S_n=a_1\left(\frac{1-r^n}{1-r}\right)$ with $r\ne 1$ .

Calculation :

Consider the sequence $2,8,32,128,\dots$ ,

It can be observed that the consecutive terms have common ratio of $\frac{8}{2}=4$ .

So, the first term is $a_1=2$ and common difference is $r=4$ .

The rule of $n^{th}$ term of a geometric series is written as $a_n=a_1{r}^{n-1}$ .

  $\begin{array}{c}{a}_n=2{\left(4\right)}^{n-1}\\ =2{\left(2^2\right)}^{n-1}\\ a_n=2^{2n-1}\end{array}$

Put 15 for $n$ to calculate $a_{15}$ .

  $\begin{array}{c}{a}_{15}=2^{2\cdot 15-1}\\ =2^{30-1}\\ =2^{29}\\ =536870912\end{array}$

The sum of first 15 terms is calculated using the formula $S_n=a_1\left(\frac{1-r^n}{1-r}\right)$ .

  $\begin{array}{c}{S}_{15}=2\left(\frac{1-4^{15}}{1-4}\right)\\ =715827882\end{array}$

Thus, the rule for $n^{th}$ term of the sequence $2,8,32,128,\dots$ is $a_n=2^{2n-1}$ , value of $a_{15}$ is $536870912$ and the sum of first 15 terms is $715827882$ .