College Algebra, Chapter 9, 9.6, Section 9.6, Problem 38

Determine the second term in the expansion of $\displaystyle \left( x^2 - \frac{1}{x} \right)^{25}$

The $r$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} $

or

$n C r - 1 (a)^{n - r + 1} (b)^{r - 1}$

In our case, $\displaystyle n = 25, a = x^2, b = \frac{-1}{x}$ and $r = 2$, so the 2nd term is


$
\begin{equation}
\begin{aligned}

=& 25 C_{2 - 1} (x^2)^{25 - 2 + 1} \left(\frac{-1}{x} \right)^{2 - 1}
\\
\\
=& 25 C_1 (x^2)^{24} \left(\frac{-1}{x} \right)^1
\\
\\
=& \frac{25!}{1! (25 - 1)!} x^{48} \left( \frac{-1}{x} \right)
\\
\\
=& -25 x^{47}


\end{aligned}
\end{equation}
$