College Algebra, Chapter 9, 9.6, Section 9.6, Problem 32

Determine the first three terms in the expansion $\displaystyle \left( x + \frac{1}{x} \right)^{40}$
Recall that the Binomial Theorem is defined as
Substituting $a = x$ and $\displaystyle b = \frac{1}{x}$ gives the first three terms are

$
\left(
\begin{array}{c}
40\\
0
\end{array}
\right)
(x)^{40},
\quad
\left(
\begin{array}{c}
40\\
1
\end{array}
\right)
(x)^{39} \left( \frac{1}{x} \right),
\quad
\left(
\begin{array}{c}
40\\
1
\end{array}
\right)
(x)^{38} \left( \frac{1}{x} \right)^2
$

From the 40th row of the Pascal's Triangle, we obtain that

$
\begin{equation}
\begin{aligned}
\left(
\begin{array}{c}
40\\
0
\end{array}
\right)
&= \frac{40!}{0!(40-0)!} = 1\\
\\
\left(
\begin{array}{c}
40\\
1
\end{array}
\right)
&= \frac{40!}{1!(40-1)!} = 40\\
\\
\left(
\begin{array}{c}
40\\
2
\end{array}
\right)
&= \frac{40!}{2!(40-2)!} = 780
\end{aligned}
\end{equation}
$

Thus, the first three terms are

$
\begin{equation}
\begin{aligned}
&= (1)(x)^{40}, \quad (40)(x)^{39} \left(\frac{1}{x}\right), \quad (780)(x)^{38} \left(\frac{1}{x}\right)^2\\
\\
&= x^{40}, \quad 40x^{38}, \quad, 780x^{36}
\end{aligned}
\end{equation}
$