Triangular numbers can also be treated as an Arithmetic progression:
[tex]1,3,6,10,15\ldots[/tex]
Where the general formula can be given by:
[tex]\begin{gathered} a_n=a_{n-1}+n \\ a_1=1 \end{gathered}[/tex]
We can check that. for n = 100, we would have:
[tex]a_{100}=a_{99}+100=a_{98}+99+100=a_{97}+97+98+100\ldots[/tex]
So, to find the 100th number, we must sum all the numbers from 1 to 100. To do so, we use the following formula:
[tex]S_n=\frac{(n+1)\times(n)}{2}[/tex]
And we find:
[tex]a_{100}=S_{100}=\frac{101\times100}{2}=5050[/tex]
The answer for 9. is 5050.
