Question

I am needing help with the second part of the problem

Answer 1

The fourth diagonal is

`1,4,10,20,35,...`

We want to find the 11th number

`\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}`

We also compute

`\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=(n(n+1))/(2) \end{gathered}`

To get the 11th number

`\begin{gathered} a_(11)=T_1+T_2+T_3+...+T_(11) \\ \\ a_(11)=\sum_{n\mathop{=}1}^(11)[(n(n+1))/(2)] \\ \\ a_(11)=286 \end{gathered}`

The answer is

`286`