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}
$
Thursday, December 25, 2014
College Algebra, Chapter 9, 9.6, Section 9.6, Problem 32
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