Chapter 9.5, Problem 41E

To determine

To find the 3^rd term in the given expansion.

Answer to Problem 41E

  $T_3=360x^3{y}^2$

Explanation of Solution

Given:

  ${\left(x-6y\right)}^5$

Concept used:

The ${\left(r+1\right)}^{\text{th}}$ term in the expansion of ${\left(a+b\right)}^n$ is,

  $T_{r+1}=C{}_r\cdot a^{n-r}\cdot b^r$

Calculation:

The 4^th term in the expansion ${\left(x-6y\right)}^5$ is,

  $\begin{array}{l}{T}_{2+1}=C{}_2\cdot x^{5-2}\cdot {\left(-6y\right)}^2\\ \Rightarrow T_3=\frac{5!}{\left(5-2\right)!\cdot 2!}{x}^3\cdot \left(36y^2\right)\\ \Rightarrow T_3=\frac{5\times 4\times 3!}{3!\cdot 2!}{x}^3\cdot \left(36y^2\right)\\ \Rightarrow T_3=\frac{20}{2}{x}^3\cdot \left(36y^2\right)\\ \Rightarrow T_3=360x^3{y}^2\end{array}$