The n-th trianglar number is given by
[tex]T_n = \dfrac{n(n+1)}2[/tex]
Then
[tex]T_4 + T_3 = \dfrac{4\times5}2 + \dfrac{3\times4}2 = \boxed{16}[/tex]
Alternatively, the triangular numbers are defined recursively by
[tex]\begin{cases}T_1 = 1 \\ T_n = T_{n-1} + n & \text{for } n>1\end{cases}[/tex]
This means
[tex]T_1=1[/tex]
[tex]T_2 = T_1 + 2 = 3[/tex]
[tex]T_3 = T_2 + 3 = 6[/tex]
[tex]T_4 = T_3 + 4 = 10[/tex]
so that
[tex]T_3+T_4=\boxed{16}[/tex]
