Chapter 12.5, Problem 42AYU

To determine

To find: The coefficient of $x^2$ in the expansion of ${\left(\sqrt{x} + \frac{3}{\sqrt{x}}\right)}^8$ using Binomial Theorem.

Answer to Problem 42AYU

Solution:

The coefficient of $x^2$ in the expansion is $5670$.

Explanation of Solution

Given:

Expression is given as ${\left(\sqrt{x} + \frac{3}{\sqrt{x}}\right)}^8$

Formula used:

The Binomial Theorem:

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

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

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 $x = \sqrt{x},a = \frac{3}{\sqrt{x}}$ and $n = 8$. The only time we will be left with an $x^2$ term is when $j=4$,

$\left(\begin{array}{c}8\\ 8 - 4\end{array}\right){\left(\frac{3}{\sqrt{x}}\right)}^{8 - 4}{\left(x\right)}^4 = \left(\begin{array}{c}8\\ 4\end{array}\right){\left(\frac{3}{\sqrt{x}}\right)}^4.x^4 = 70 \times \frac{81}{x^2}{x}^4 = 5670x^2$

The coefficient of $x^2$ in the expansion is $5670$.