Chapter 9, Problem 2CT

To determine

To calculate: The expression for the $n\text{th}$ term of the sequence $\frac{3}{1!},\frac{4}{2!},\frac{5}{3!},\frac{6}{4!},\frac{7}{5!},\dots$ .

Answer to Problem 2CT

The expression for the $n\text{th}$ term of the sequence $\frac{3}{1!},\frac{4}{2!},\frac{5}{3!},\frac{6}{4!},\frac{7}{5!},\dots$ is $a_n=\frac{n+2}{n!}$ .

Explanation of Solution

Given information:

The sequence $\frac{3}{1!},\frac{4}{2!},\frac{5}{3!},\frac{6}{4!},\frac{7}{5!},\dots$ .

Formula used:

Terms of a sequence always follow a pattern. Express the first second third and similarly the remaining terms of sequence in a similar pattern to express the apparent $nth$ term of sequence.

Calculation:

Consider the sequence $\frac{3}{1!},\frac{4}{2!},\frac{5}{3!},\frac{6}{4!},\frac{7}{5!},\dots$ .

Denote $a_n=\frac{3}{1!},\frac{4}{2!},\frac{5}{3!},\frac{6}{4!},\frac{7}{5!},\dots$

Observe that first term denoted by $a_1$ is $\frac{3}{1!}$ .

First term can be expressed as,

  $a_1=\frac{1+2}{1!}$

The second term denoted by $a_2$ is $\frac{4}{2!}$ .

Second term can be expressed,

  $a_2=\frac{2+2}{2!}$

Similarly, third term can be expressed,

  $a_3=\frac{3+2}{3!}$

Similarly, fourth term can be expressed,

  $a_4=\frac{4+2}{4!}$

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

So, apparent term is $\left(n+2\right)$ divided by $n!$ where n starts from 1.

Mathematically, it is expressed as, $a_n=\frac{n+2}{n!}$

Thus, the expression for the $n\text{th}$ term of the sequence $\frac{3}{1!},\frac{4}{2!},\frac{5}{3!},\frac{6}{4!},\frac{7}{5!},\dots$ is $a_n=\frac{n+2}{n!}$ .