Solution
The fourth diagonal is
[tex]1,4,10,20,35,...[/tex]
We want to find the 11th number
[tex]\begin{gathered} T_1=a_1=1 \\ T_2=a_2-a_1=4-1=3 \\ T_3=a_3-a_2=10-4=6 \\ T_4=a_4-a_3=20-10=10 \\ T_5=a_5-a_4=35-20=15 \end{gathered}[/tex]
We also compute
[tex]\begin{gathered} T_1=1 \\ T_2-T_1=3-1=2 \\ T_3-T_2=6-3=3 \\ T_4-T_3=10-6=4 \\ . \\ . \\ . \\ T_n-T_{n-1}=n \\ Adding\text{ up the above we have} \\ T_n=1+2+3+...+n \\ T_n=\frac{n(n+1)}{2} \end{gathered}[/tex]
To get the 11th number
[tex]\begin{gathered} a_{11}=T_1+T_2+T_3+...+T_{11} \\ \\ a_{11}=\sum_{n\mathop{=}1}^{11}[\frac{n(n+1)}{2}] \\ \\ a_{11}=286 \end{gathered}[/tex]
The answer is
[tex]286[/tex]
