Chapter 12.3, Problem 24E

To determine

Calculate the first five terms of the given sequence $a_n={\left(-1\right)}^n{\left(2\right)}^n$ and find whether it is geometric. If it is geometric then also calculate the common ratio, and express the $n^{th}$ term of the sequence in the standard form $a_n=a{r}^{n-1}$ .

Answer to Problem 24E

The first five term of the given sequence $a_n={\left(-1\right)}^n{\left(2\right)}^n$ is $-2,4,-8,16\&-32$

The given sequence in geometric

Having the common ratio, $r=-2$

The $n^{th}$ term of the sequence in the standard form $a_n=a{r}^{n-1}$ is $a_n=-2*{\left(-2\right)}^{n-1}$

Explanation of Solution

Given information:

It is given in the question that,

the given sequence is $a_n={\left(-1\right)}^n{\left(2\right)}^n$

Formula used:

  $a_n=a{r}^{n-1}$ & $r\Rightarrow \frac{a_n}{a_{n-1}}=\frac{a_{n+1}}{a_n}$

Where,

  $\begin{array}{l}aisthe{1}^{st}term\\ nisthenumberofterm\\ risthecommonratio\end{array}$

Calculation:

For finding the first five term put $n=1,2,3,4,5$ in the given sequence is $a_n={\left(-1\right)}^n{\left(2\right)}^n$

So,

By putting $n=1$ in $a_n={\left(-1\right)}^n{\left(2\right)}^n$

  $\begin{array}{l}{a}_1={\left(-1\right)}^1{\left(2\right)}^1\\ =-1*2\\ =-2\end{array}$

By putting $n=2$ in $a_n={\left(-1\right)}^n{\left(2\right)}^n$

  $\begin{array}{l}{a}_2={\left(-1\right)}^2{\left(2\right)}^2\\ =1*4\\ =4\end{array}$

By putting $n=3$ in $a_n={\left(-1\right)}^n{\left(2\right)}^n$

  $\begin{array}{l}{a}_3={\left(-1\right)}^3{\left(2\right)}^3\\ =-1*8\\ =-8\end{array}$

By putting $n=4$ in $a_n={\left(-1\right)}^n{\left(2\right)}^n$

  $\begin{array}{l}{a}_4={\left(-1\right)}^4{\left(2\right)}^4\\ =1*16\\ =16\end{array}$

By putting $n=5$ in $a_n={\left(-1\right)}^n{\left(2\right)}^n$

  $\begin{array}{l}{a}_5={\left(-1\right)}^5{\left(2\right)}^5\\ =-1*32\\ =-32\end{array}$

By using the formula $r\Rightarrow \frac{a_n}{a_{n-1}}=\frac{a_{n+1}}{a_n}$

  $\begin{array}{l}r\Rightarrow \frac{a_n}{a_{n-1}}=\frac{a_{n+1}}{a_n}\\ \Rightarrow \frac{4}{-2}=\frac{-8}{4}\\ \Rightarrow -2=-2\end{array}$

Here, the sequence satisfied the given formula concept so the given sequence $a_n={\left(-1\right)}^n{\left(2\right)}^n$ which shows that the common ratio of two consecutive term is same.

So, the given sequence in geometric

Having the common ratio, $r=-2$

The $n^{th}$ term of the sequence in the standard form $a_n=a{r}^{n-1}$ is $a_n=-2*{\left(-2\right)}^{n-1}$