Chapter 11.7, Problem 12WE

To determine

To calculate:

Expansion and simplification of expression

  ${\left(a+\frac{1}{a}\right)}^5$

Answer to Problem 12WE

The simplified expression is $a^5+5a^3+10a+\frac{10}{a}+\frac{5}{a^3}+\frac{1}{a^5}$

Explanation of Solution

Given information:

The given expression is,

  ${\left(a+\frac{1}{a}\right)}^5$

Formula:

The formula of binomial theorem is given by,

  ${\left(a+b\right)}^n={\displaystyle {\sum }_{i=0}^n{\left(\right)}_{b^i}}$

Calculation:

The given expression is,

  ${\left(a+\frac{1}{a}\right)}^5$

Here, consider

  $\begin{array}{l}a=a\\ b=\frac{1}{a}\end{array}$

Now, expanding the given expression using Binomial theorem,

  $\begin{array}{l}=\frac{5!}{0!\left(5-0\right)!}{a}^5{\left(\frac{1}{a}\right)}^0+\frac{5!}{1!\left(5-1\right)!}{a}^4{\left(\frac{1}{a}\right)}^1+\frac{5!}{2!\left(5-2\right)!}{a}^3{\left(\frac{1}{a}\right)}^2\\ \frac{5!}{3!\left(5-3\right)!}{a}^2{\left(\frac{1}{a}\right)}^3+\frac{5!}{4!\left(5-4\right)!}{a}^1{\left(\frac{1}{a}\right)}^4+\frac{5!}{5!\left(5-5\right)!}{a}^0{\left(\frac{1}{a}\right)}^5\end{array}$

Simplifying the above expression,

  $a^5+5a^3+10a+\frac{10}{a}+\frac{5}{a^3}+\frac{1}{a^5}$