Consider this triangular set of numbers. Determine the last number in the $99$th row.
I came up with the formula for the $n$'th term, and got $\frac{1}{2}n^{2} +\frac{1}{2}n$
The issue is that the last term in the $99$th row is not the last term. So the $3$rd overall term is the last term of the second row, the 6th overall term is the last term in the third row, and the $10$th overall term is the last in the fourth row, etc.
So the pattern is $3,6,10, \dots$ and the formula for $a_{n} = \frac{1}{2}n^{2}+1.5n+1$
My issue is, how can I find out what term number the last number of the $99$th term is? that way I can find out what $n$ is to plug into my original equation
Hint. The last term of the $r$-th row is $n$-th triangular number where $$n=1+2+3+\dots+r.$$