Chapter 12.5, Problem 21AYU

In Problems 17-28, expand each expression using the Binomial Theorem.

${\left(3x+1\right)}^4$

To determine

To find: The expansion of the given binomial expression ${\left(3x + 1\right)}^4$ using Binomial Theorem.

Answer to Problem 21AYU

Solution:

The expansion of the expression is $81x^4 + 108x^3 + 54x^2 + 12x + 1$.

Explanation of Solution

Given:

Expression is given as ${\left(3x + 1\right)}^4$.

Formula used:

The Binomial Theorem:

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

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

Here $x = 3x$, $a = 1$ and $n = 4$. Applying Binomial theorem,

$\begin{array}{l}{\left(3x + 1\right)}^4 = \left(\begin{array}{c}4\\ 0\end{array}\right){\left(3x\right)}^4 + \left(\begin{array}{c}4\\ 1\end{array}\right)\left(1\right){\left(3x\right)}^3 + \left(\begin{array}{c}4\\ 2\end{array}\right){\left(1\right)}^2{\left(3x\right)}^2 + \left(\begin{array}{c}4\\ 3\end{array}\right){\left(1\right)}^3{\left(3x\right)}^1 + \left(\begin{array}{c}4\\ 4\end{array}\right){\left(1\right)}^4\\ = 1.81x^4 + 4.\left(1\right)27x^3 + 6.\left(1\right)9x^2 + 4.\left(1\right)3x^1 + 1.\left(1\right)\\ {\left(3x + 1\right)}^4 = 81x^4 + 108x^3 + 54x^2 + 12x + 1\end{array}$