Answer
$a_{1}=1$
$a_{2}= 3$
$a_{3}=6$
$a_{4}=10$
$a_{10}=55$
Work Step by Step
We are given:
$a_{n}=\left(\begin{array}{l}n+1\\2\end{array}\right)$
We evaluate:
$a_{1}=\left(\begin{array}{l}1+1\\2\end{array}\right)=\frac{2!}{2!0!}=1$
$a_{2}= \left(\begin{array}{l}2+1\\2\end{array}\right)=\frac{3!}{2!1!}=3$
$a_{3}= \left(\begin{array}{l}3+1\\2\end{array}\right)=\frac{4!}{2!2!}=6$
$a_{4}= \left(\begin{array}{l}4+1\\2\end{array}\right)=\frac{5!}{2!3!}=10$
$a_{10}= \left(\begin{array}{l}10+1\\2\end{array}\right)=\frac{11!}{2!9!}=55$
