College Algebra, Chapter 9, Review Exercises, Section Review Exercises, Problem 80

Determine the first three terms in the expression of $\displaystyle \left( b^{\frac{-2}{3}} + b^{\frac{1}{3}} \right)^{20}$
The $n$th term of the binomial expansion $(a+b)^n$ is defined as
$\displaystyle \left( \begin{array}{c} n \\ r -1 \end{array} \right) (a)^{n-r+1} (b)^{r-1} \qquad \text{or} \qquad nC_{r-1} = (a)^{n-r+1} (b)^{r-1}$

If we rewrite the expression as $\displaystyle \left( b^{\frac{1}{3}} + b^{\frac{-2}{3}} \right)^{20}$ we have, $n = 20$, $a = b^{\frac{1}{3}}$ and $b = b^{\frac{-2}{3}}$. So the first term is


$\begin{equation} \begin{aligned} &= \left( \begin{array}{c} 20 \\ 1-1 \end{array} \right) \left( b^{\frac{1}{3}} \right)^{20 - 1 +1} \left( b^{\frac{-2}{3}} \right)^{1-1} \\ \\ &= \left( \begin{array}{c} 20 \\ 0 \end{array} \right) \left( b^{\frac{1}{3}} \right)^{20} \left( b^{\frac{-2}{3}} \right)^{0} \\ \\ &= b^{\frac{20}{3}} \end{aligned} \end{equation}$


The second term is

$\begin{equation} \begin{aligned} &= \left( \begin{array}{c} 20 \\ 2-1 \end{array} \right) \left( b^{\frac{1}{3}} \right)^{20 - 2 +1} \left( b^{\frac{-2}{3}} \right)^{2-1} \\ \\ &= \left( \begin{array}{c} 20 \\ 1 \end{array} \right) \left( b^{\frac{1}{3}} \right)^{19} \left( b^{\frac{-2}{3}} \right) \\ \\ &= \left( \begin{array}{c} 20 \\ 1 \end{array} \right) \left( b^{\frac{19}{3}} \right) \left( b^{\frac{-2}{3}} \right) \\ \\ &= \left( \begin{array}{c} 20 \\ 1 \end{array} \right) (b)^{\frac{19}{3} - \frac{2}{3}} \\ \\ &= 20b^{\frac{17}{3}} \end{aligned} \end{equation}$


Then, the third term is

$\begin{equation} \begin{aligned} &= \left( \begin{array}{c} 20 \\ 3-1 \end{array} \right) \left( b^{\frac{1}{3}} \right)^{20 - 3 +1} \left( b^{\frac{-2}{3}} \right)^{3-1} \\ \\ &= \left( \begin{array}{c} 20 \\ 2 \end{array} \right) \left( b^{\frac{1}{3}} \right)^{18} \left( b^{\frac{-2}{3}} \right)^{2} \\ \\ &= \left( \begin{array}{c} 20 \\ 2 \end{array} \right) \left( b^{\frac{18}{3}} \right) \left( b^{\frac{-4}{3}} \right) \\ \\ &= \left( \begin{array}{c} 20 \\ 2 \end{array} \right) \left( b^{\frac{18}{3}- \frac{4}{3}} \right) \\ \\ &= 190b^{\frac{14}{3}} \end{aligned} \end{equation}$

Therefore, the first three terms are,
$b^{\frac{20}{3}} + 20b^{\frac{17}{3}}$ and $190b^{\frac{14}{3}}$ respectively.