Answer
1, 3, 6, 10,
$a_{10}=55$
Work Step by Step
$\displaystyle \left(\begin{array}{l}
n\\
r
\end{array}\right)=\frac{n!}{r!(n-r)!}$
$n!=n(n-1)(n-2)\cdot...\cdot 2\cdot 1$
$0!=1$
----------------
$\displaystyle \left(\begin{array}{l}
n+1\\
2
\end{array}\right)=\frac{(n+1)!}{2!(n+1-2)!}=\frac{(n+1)!}{2!(n-1)!}$
=$\displaystyle \frac{(n+1)n\cdot(n-1)!}{2\cdot 1\cdot(n-1)!}=\frac{(n+1)n}{2}$
$a_{1}= \left(\begin{array}{l}
1+1\\
2
\end{array}\right) =\displaystyle \frac{2(1)}{2}=1$,
$a_{2}= \left(\begin{array}{l}
2+1\\
2
\end{array}\right) =\displaystyle \frac{3(2)}{2}=3$,
$a_{3}= \left(\begin{array}{l}
3+1\\
2
\end{array}\right) =\displaystyle \frac{4(3)}{2}=6$,
$a_{4}=\left(\begin{array}{l}
4+1\\
2
\end{array}\right) =\displaystyle \frac{5(4)}{2}=10$,
$a_{10}=\left(\begin{array}{l}
10+1\\
2
\end{array}\right) =\displaystyle \frac{11(10)}{2}=55$
