Chapter 11.3, Problem 32WE

To determine

To calculate:

The following sequence is arithmetic, geometric or neither and the formula for $n^{th}$ term:

  $e^x,e^{2x},e^{3x},e^{4x},\mathrm{...}$

Answer to Problem 32WE

The sequence is geometric sequence and formula for $n^{th}$ term is $a_n=e^{xn}$

Explanation of Solution

Given information:

  $e^x,e^{2x},e^{3x},e^{4x},\mathrm{...}$

Calculation:

Check the common difference first to categorize the given sequence into arithmetic, geometric or neither:

  $\begin{array}{l}{a}_1=e^x\\ a_2=e^{2x}\\ a_3=e^{3x}\\ a_4=e^{4x}\end{array}$

Now, the common difference is calculated as:

  $\begin{array}{l}d=a_2-a_1=e^{2x}-e^x\\ d=a_3-a_2=e^{2x}-e^x-e^x=e^{2x}-2e^x\\ d=a_4-a_3=e^{2x}-2e^x-e^x=e^{2x}-3e^x\end{array}$

Therefore, the given sequence is not arithmetic sequence as the common ratio is not same.

Now, check for geometric sequence:

  $\begin{array}{l}r=\frac{a_2}{a_1}=\frac{e^{2x}}{e^x}=e^x\\ r=\frac{a_3}{a_2}=\frac{e^{3x}}{e^{2x}}=e^x\\ r=\frac{a_4}{a_3}=\frac{e^{4x}}{e^{3x}}=e^x\end{array}$

Therefore, the given sequence is geometric sequence.

Now, the formula for $n^{th}$ term is calculated as:

  $\begin{array}{l}{a}_n=a{r}^{\left(n-1\right)}\\ a_n=e^x{\left(e^x\right)}^{\left(n-1\right)}\\ a_n=e^x\cdot e^{x\left(n-1\right)}\\ a_n=e^x{}^{n-x+x}\\ a_n=e^{xn}\end{array}$