Chapter 14, Problem 1MCP

To determine

To Write:

First five terms of the sequence $a_n={\left(-1\right)}^{n+1}\frac{n}{\left(n+1\right)!}$ .

Answer to Problem 1MCP

Solution:

The first five terms of the sequence are: $\frac{1}{2},-\frac{1}{3},\frac{1}{8},-\frac{1}{30},\frac{1}{144}$.

Explanation of Solution

Given:

The sequence $a_n={\left(-1\right)}^{n+1}\frac{n}{\left(n+1\right)!}$ .

Calculation:

  $a_n={\left(-1\right)}^{n+1}\frac{n}{\left(n+1\right)!}$

For $n=1$

, the first term is:

  $\begin{array}{c}{a}_1={\left(-1\right)}^{1+1}\frac{1}{\left(1+1\right)!}\\ ={\left(-1\right)}^2\frac{1}{\left(2\right)!}\\ =\left(1\right)\frac{1}{2\times 1}\\ =\frac{1}{2}\end{array}$

For $n=2$ , the second term is:

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

For $n=3$ , the third term is:

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

For $n=4$ , the fourth term is:

  $\begin{array}{c}{a}_4={\left(-1\right)}^{4+1}\frac{4}{\left(4+1\right)!}\\ ={\left(-1\right)}^5\frac{4}{\left(5\right)!}\\ =\left(-1\right)\frac{4}{5\times 4\times 3\times 2\times 1}\\ =-\frac{1}{30}\end{array}$

For $n=5$ , the fifth term is:

  $\begin{array}{c}{a}_5={\left(-1\right)}^{5+1}\frac{5}{\left(5+1\right)!}\\ ={\left(-1\right)}^6\frac{5}{\left(6\right)!}\\ =\left(1\right)\frac{5}{6\times 5\times 4\times 3\times 2\times 1}\\ =\frac{1}{144}\end{array}$

Therefore, the first five terms of the sequence are: $\frac{1}{2},-\frac{1}{3},\frac{1}{8},-\frac{1}{30},\frac{1}{144}$ .

Conclusion:

The first five terms of the sequence are: $\frac{1}{2},-\frac{1}{3},\frac{1}{8},-\frac{1}{30},\frac{1}{144}$ .