Chapter 12.5, Problem 40AYU

To determine

To find: The coefficient of $x^0$ in the expansion of ${\left(x - \frac{1}{x^2}\right)}^9$ using Binomial Theorem.

Answer to Problem 40AYU

Solution:

The coefficient of $x^0$ in the expansion is $- 84$.

Explanation of Solution

Given:

Expression is given as $$ {\left(x - \frac{1}{x^2}\right)}^9$$

Formula used:

The Binomial Theorem:

Let $x$ and $a$ be real numbers. For any positive integer $n$, we have

$\begin{array}{l}{\left(x + a\right)}^n = \left(\begin{array}{c}n\\ 0\end{array}\right)x^n + \left(\begin{array}{c}n\\ 1\end{array}\right)a{x}^{n - 1} + \cdots + \left(\begin{array}{c}n\\ j\end{array}\right)a^j{x}^{n - j} + \cdots + \left(\begin{array}{c}n\\ n\end{array}\right)a^n\\ {\left(x + a\right)}^n = {\displaystyle \sum _{k = 0}^n\left(\begin{array}{c}n\\ k\end{array}\right)x^{n - j}{a}^j}\end{array}$

The binomial theorem can be used to find a particular term in an expansion without writing the entire expansion.

Based on the expansion of ${\left(x + a\right)}^n$, the term containing $x^j$ is $\left(\begin{array}{c}n\\ n - j\end{array}\right)a^{n - j}{x}^j$.

Here $x = x,a = - \frac{1}{x^2}$ and $n = 9$. The only time we will be left with an $x^0$ term is when $j = 6$, $x^0$ is $\left(\begin{array}{c}9\\ 9 - 6\end{array}\right){\left( - \frac{1}{x^2}\right)}^{9 - 6}{\left(x\right)}^6 = \left(\begin{array}{c}9\\ 3\end{array}\right){\left( - \frac{1}{x^2}\right)}^3\cdot {\left(x\right)}^6 = 84 \times - 1.1 = - 84$.

The coefficient of $x^0$ in the expansion is $- 84$.