Chapter 11.8, Problem 24WE

To determine

To calculate:

Simplify the specified term

Answer to Problem 24WE

The term containing $b^9$ in ${\left(a+3b^3\right)}^5$ is $4^{th}$ term .

Explanation of Solution

Given information:

Expression:

  ${\left(a+3b^3\right)}^5$

Term:

  $b^9$

Formula used:

Binomial theorem:

  ${\left(a+b\right)}^n={\displaystyle {\sum }_{i=0}^n\left(\begin{array}{c}n\\ i\end{array}\right)}{a}^{\left(n-i\right)}{b}^i$

Where,

  $a=$ first term

  $b=$ second term

Calculation:

Consider expression ${\left(a+3b^3\right)}^5$

Suppose, the term $t_{r+1}$ contains $b^9$

Apply the binomial theorem for ${\left(a+3b^3\right)}^5$

  ${\left(a+b\right)}^n={\displaystyle {\sum }_{i=0}^n\left(\begin{array}{c}n\\ i\end{array}\right)}{a}^{\left(n-i\right)}{b}^i$

Where,

  $a=a,b=3b^3,n=5$

Therefore, with the help of the above theorem, $t_{r+1}$ can be calculated as $t_{r+1}=\left(\begin{array}{c}n\\ r\end{array}\right)a^{n-r}{b}^r$

Now, in the given binomial ${\left(a+3b^3\right)}^5$

Therefore, to find the term that contains $b^9$ ,

So, first find the value of $r$

Put the values of $a,b,$ and $n$ in $t_{r+1}=\left(\begin{array}{c}n\\ r\end{array}\right)a^{n-r}{b}^r$ ,

  $\begin{array}{l}{t}_{r+1}=\left(\begin{array}{c}n\\ r\end{array}\right)a^{n-r}{b}^r\\ =\left(\begin{array}{c}5\\ r\end{array}\right)a^{5-r}{\left(3b^3\right)}^r\\ =3\left(\begin{array}{c}5\\ r\end{array}\right)a^{5-r}\left(b^{3r}\right)\end{array}$

Thus $t_{r+1}$ contains $b^9$ if

  $\begin{array}{l}3r=9\\ r=\frac{9}{3}\\ r=3\end{array}$

Therefore,

  $\begin{array}{l}{t}_{r+1}=t_{3+1}\\ =t_4\end{array}$

Hence, $4^{th}$ term contains $b^9$ in the expansion of ${\left(a+3b^3\right)}^5$ .