College Algebra, Chapter 9, Review Exercises, Section Review Exercises, Problem 6

Determine the first four terms and the tenth term of the sequence $\displaystyle a_n =
\left(
\begin{array}{c}
n + 1 \\
2
\end{array}
\right)
$



$
\displaystyle
a_1 =
\left(
\begin{array}{c}
1 + 1 \\
2
\end{array}
\right)
=
\left(
\begin{array}{c}
2 \\
2
\end{array}
\right)
=
\frac{2!}{2!(2-2)^2} = 1
$



$
\displaystyle
a_2 =
\left(
\begin{array}{c}
2 + 1 \\
2
\end{array}
\right)
=
\left(
\begin{array}{c}
3 \\
2
\end{array}
\right)
=
\frac{3!}{2!(3-2)^2} = 3
$



$
\displaystyle
a_3 =
\left(
\begin{array}{c}
3 + 1 \\
2
\end{array}
\right)
=
\left(
\begin{array}{c}
4 \\
2
\end{array}
\right)
=
\frac{4!}{2!(4-2)^2} = 6
$



$
\displaystyle
a_4 =
\left(
\begin{array}{c}
4 + 1 \\
2
\end{array}
\right)
=
\left(
\begin{array}{c}
5 \\
2
\end{array}
\right)
=
\frac{5!}{2!(5-2)^2} = 10
$

$\cdot$
$\cdot$
$\cdot$


$
\displaystyle
a_{10} =
\left(
\begin{array}{c}
10 + 1 \\
2
\end{array}
\right)
=
\left(
\begin{array}{c}
11 \\
2
\end{array}
\right)
=
\frac{11!}{2!(11-2)^2} = 55
$