College Algebra, Chapter 9, 9.6, Section 9.6, Problem 30

Determine the first four terms in the expansion $\displaystyle \left( x^{\frac{1}{2}} + 1 \right)^{30}$
Recall that the Binomial Theorem is defined as
Substitute $ a = x^{\frac{1}{2}}$ and $b = 1$ gives the first four terms as,

$
=
\left(
\begin{array}{c}
30\\
0
\end{array}
\right)
\left( x^{\frac{1}{2}} \right)^{30} +
\left(
\begin{array}{c}
30\\
1
\end{array}
\right)
\left( x^{\frac{1}{2}} \right)^{29}(1) +
\left(
\begin{array}{c}
30\\
2
\end{array}
\right)
\left( x^{\frac{1}{2}} \right)^{28} (1)^2 +
\left(
\begin{array}{c}
30\\
3
\end{array}
\right)
\left( x^{\frac{1}{2}} \right)^{27} (1)^3
$

From the 30th row of the Pascal's Triangle,

$
\begin{equation}
\begin{aligned}
\left(
\begin{array}{c}
30\\
0
\end{array}
\right)
&= \frac{30!}{0!(30-0)!} = 1\\
\\
\left(
\begin{array}{c}
30\\
1
\end{array}
\right)
&= \frac{30!}{1!(30-1)!} = 30\\
\\
\left(
\begin{array}{c}
30\\
2
\end{array}
\right)
&= \frac{30!}{2!(30-2)!} = 435\\
\\
\left(
\begin{array}{c}
30\\
3
\end{array}
\right)
&= \frac{30!}{3!(30-3)!} = 4060\\
\\
\end{aligned}
\end{equation}
$

Thus, the first four terms are

$
\begin{equation}
\begin{aligned}
&= (1) \left( x^{\frac{1}{2}} \right)^{30}, \quad
(30)\left( x^{\frac{1}{2}} \right)^{29}(1), \quad
(435)\left( x^{\frac{1}{2}} \right)^{28}(1)^2, \quad
\text{ and }
\left( 4060 x^{\frac{1}{2}} \right)^{27} (1)^3\\
\\
&= x^{15}, \quad
30x^{\frac{29}{2}}, \quad
435x^{14}, \quad
4060x^{\frac{27}{2}}
\end{aligned}
\end{equation}
$