Chapter 12.6, Problem 34E

To determine

To find:Fifth term in the expansion of ${\left(ab-1\right)}^{20}$ using Binomial Theorem.

Answer to Problem 34E

Fifth term = $a_5=4845{\left(ab\right)}^{16}$

Explanation of Solution

Given information:

  ${\left(ab-1\right)}^{20}$

Calculation:

According to the Binomial theorem, expansion of ${\left(x+y\right)}^n$ with non-negative integer $n$ is:

  ${\left(x+y\right)}^n=\left(\begin{array}{l}n\\ 0\end{array}\right)x^n{y}^0+\left(\begin{array}{l}n\\ 1\end{array}\right)x^{n-1}{y}^1+\mathrm{.............}+\left(\begin{array}{l}n\\ n-1\end{array}\right)x^1{y}^{n-1}+\left(\begin{array}{l}n\\ n\end{array}\right)x^0{y}^n$

Where $n\ge 0$ , and $\left(\begin{array}{l}n\\ k\end{array}\right)$ denotes the binomial coefficient for $k=0,1,\mathrm{2........}n-1,n$

The ${\left(r+1\right)}^{th}$ term of the binomial expansion of ${\left(x+y\right)}^n$ is $a_{r+1}=\left(\begin{array}{l}n\\ r\end{array}\right)x^{n-r}{y}^r$

We know that,

  $\left(\begin{array}{l}n\\ k\end{array}\right)=\frac{n!}{\left(n-k\right)!k!}$

Given

  ${\left(ab-1\right)}^{20}$

Here $n=20$

We need to find fifth term

Therefore

  $r+1=5$

  $\Rightarrow r=4$

Substitute $r=4$

We get

  $\begin{array}{l}{a}_{4+1}=\left(\begin{array}{l}20\\ 4\end{array}\right){\left(ab\right)}^{20-4}{\left(-1\right)}^4\\ \Rightarrow a_5=\left(\begin{array}{l}20\\ 4\end{array}\right){\left(ab\right)}^{16}\\ \left(\begin{array}{l}20\\ 4\end{array}\right)=\frac{20!}{\left(20-4\right)!4!}=\frac{20!}{16!4!}=\frac{20\times 19\times 18\times 17}{4\times 3\times 2\times 1}=4845\\ \Rightarrow a_5=4845{\left(ab\right)}^{16}\end{array}$

Therefore

Fifth term of expansion of ${\left(ab-1\right)}^{20}$ is

  $a_5=4845{\left(ab\right)}^{16}$