Chapter 11.6, Problem 32PPS

To determine

The expansion of the binomial ${\left(3c+\frac{1}{3}\right)}^5$

Answer to Problem 32PPS

The expand form is:

  $243c^5+135c^4+30c^3+\frac{10c^2}{3}+\frac{5c}{27}+\frac{1}{243}$

Explanation of Solution

Given information:

The given expression is as:

  ${\left(3c+\frac{1}{3}\right)}^5$

Concept used:

1. The binomial expansion for ${\left(x+y\right)}^n$ is as:

  ${\left(x+y\right)}^n={\displaystyle \sum _{r=0}^n\left(\begin{array}{c}n\\ r\end{array}\right)x^{n-r}{y}^r}$

Where the coefficient for term $x^{n-r}{y}^r$ is $\left(\begin{array}{c}n\\ r\end{array}\right)=\frac{n!}{r!\left(n-r\right)!}$

Calculation:

Since, the given expression is as:

  ${\left(3c+\frac{1}{3}\right)}^5$

On comparing this with ${\left(x+y\right)}^n$ , the value of $n$ is 5

  $\Rightarrow n=5$

On employing binomial expansion, we can express given expression as

  ${\left(3c+\frac{1}{3}\right)}^5={\displaystyle \sum _{r=0}^5\left(\begin{array}{c}5\\ r\end{array}\right){\left(3c\right)}^{5-r}{\left(\frac{1}{3}\right)}^r}$

Therefore, the expand for ${\left(3c+\frac{1}{3}\right)}^5$ is

  $\begin{array}{c}{\left(3c+\frac{1}{3}\right)}^5={\displaystyle \sum _{r=0}^5\left(\begin{array}{c}5\\ r\end{array}\right){\left(3c\right)}^{5-r}{\left(\frac{1}{3}\right)}^r}\\ =\frac{5!}{\left(5-0\right)!0!}{\left(3c\right)}^{5-0}\cdot {\left(\frac{1}{3}\right)}^0+\frac{5!}{\left(5-1\right)!1!}{\left(3c\right)}^{5-1}{\left(\frac{1}{3}\right)}^1+\frac{5!}{\left(5-2\right)!2!}{\left(3c\right)}^{5-2}\cdot {\left(\frac{1}{3}\right)}^2\\ +\frac{5!}{\left(5-3\right)!3!}{\left(3c\right)}^{5-3}\cdot {\left(\frac{1}{3}\right)}^3+\frac{5!}{\left(5-4\right)!4!}{\left(3c\right)}^{5-4}\cdot {\left(\frac{1}{3}\right)}^4+\frac{5!}{\left(5-5\right)!5!}{\left(3c\right)}^{5-5}\cdot {\left(\frac{1}{3}\right)}^5\\ =243c^5+135c^4+30c^3+\frac{10c^2}{3}+\frac{5c}{27}+\frac{1}{243}\end{array}$

Thus, the expand of the given binomial is,

  $243c^5+135c^4+30c^3+\frac{10c^2}{3}+\frac{5c}{27}+\frac{1}{243}$