Chapter 12.5, Problem 32AYU

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

The coefficient of $x^3$ in the expansion of ${\left(2x+1\right)}^{12}$

To determine

To find: The coefficient of $x^3$ in the expansion of ${\left(2x + 1\right)}^{12}$ using Binomial Theorem.

Answer to Problem 32AYU

Solution:

The coefficient of $x^3$ in the expansion is $1760$.

Explanation of Solution

Given:

Expression is given as ${\left(2x + 1\right)}^{12}$

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 $x = 2x, a = 1$ and $n = 12$. Applying Binomial theorem,

$x^3$ is $\left(\begin{array}{c}12\\ 12 - 3\end{array}\right){\left(1\right)}^{12 - 3}{\left(2x\right)}^3 = \left(\begin{array}{c}12\\ 9\end{array}\right){\left(1\right)}^9.8.x^3 = 220 \times 8x^3 = 1760x^7$

The coefficient of $x^3$ in the expansion is $1760$.