Chapter 9.1, Problem 50E

To determine

To calculate: The expression for the $n\text{th}$ term of the sequence $\frac{2}{1},\frac{6}{3},\frac{24}{7},\frac{120}{15},\frac{720}{31},\dots$ .

Answer to Problem 50E

The expression for the $n\text{th}$ term of the sequence $\frac{2}{1},\frac{6}{3},\frac{24}{7},\frac{120}{15},\frac{720}{31},\dots$ is $a_n=\frac{\left(n+1\right)!}{2^n-1}$ .

Explanation of Solution

Given information:

The sequence $\frac{2}{1},\frac{6}{3},\frac{24}{7},\frac{120}{15},\frac{720}{31},\dots$ .

Formula used:

Recursive formula for sequence is when the previous term is used to evaluate the next term of the sequence.

Calculation:

Consider the sequence $\frac{2}{1},\frac{6}{3},\frac{24}{7},\frac{120}{15},\frac{720}{31},\dots$ .

Denote $a_n=\frac{2}{1},\frac{6}{3},\frac{24}{7},\frac{120}{15},\frac{720}{31},\dots$

Observe that first term denoted by $a_1$ is $\frac{2}{1}$ .

Rewrite it as, $\frac{2}{1}=\frac{\left(1+1\right)!}{2^0-1}$

The second term denoted by $a_2$ is $\frac{6}{3}$ .

Second term can be expressed,

  $\begin{array}{c}{a}_2=\frac{\left(2+1\right)!}{2^2-1}\\ =\frac{3!}{4-1}\\ =\frac{6}{3}\end{array}$

Similarly, third term can be expressed,

  $\begin{array}{c}{a}_3=\frac{\left(3+1\right)!}{2^3-1}\\ =\frac{4!}{8-1}\\ =\frac{24}{7}\end{array}$

Similarly, fourth term can be expressed,

  $\begin{array}{c}{a}_4=\frac{\left(4+1\right)!}{2^4-1}\\ =\frac{5!}{16-1}\\ =\frac{120}{15}\end{array}$

The pattern continues for the rest of the terms of the sequence.

So, apparent term is $\left(n+1\right)!$ divided by $2^n-1$ .

Mathematically, it is expressed as, $a_n=\frac{\left(n+1\right)!}{2^n-1}$

Thus, the expression for the $n\text{th}$ term of the sequence $\frac{2}{1},\frac{6}{3},\frac{24}{7},\frac{120}{15},\frac{720}{31},\dots$ is $a_n=\frac{\left(n+1\right)!}{2^n-1}$ .