Chapter 12.5, Problem 30AYU

In Problems 29-42, use the Binomial Theorem to find the indicated coefficient or term.

The coefficient of $x^3$ in the expansion of ${\left(x-3\right)}^{10}$

To determine

To find: The coefficient of $x^3$ in the expansion of ${\left(x - 3\right)}^{10}$ using Binomial Theorem.

Answer to Problem 30AYU

Solution:

The coefficient of $x^3$ in the expansion is $-262440$.

Explanation of Solution

Given:

Expression is given as ${\left(x - 3\right)}^{10}$

Formula used:

The Binomial Theorem:

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

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

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

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 $a = -3$ and $n = 10$. Applying Binomial theorem,

$x^3$ is $\left(\begin{array}{c}10\\ 10 - 3\end{array}\right){\left(-3\right)}^{10 - 3}{x}^3 = \left(\begin{array}{c}10\\ 7\end{array}\right){\left(-3\right)}^7{x}^3 = 210 \times -2187x^3 = -262440x^3$

The coefficient of $x^3$ in the expansion is $-262440$.