Question

Here’s the problem (Sorry I thought it will be easier if I just sent a picture):

Answer 1

Triangular numbers can also be treated as an Arithmetic progression:

`1,3,6,10,15\ldots`

Where the general formula can be given by:

`\begin{gathered} a_n=a_(n-1)+n \\ a_1=1 \end{gathered}`

We can check that. for n = 100, we would have:

`a_(100)=a_(99)+100=a_(98)+99+100=a_(97)+97+98+100\ldots`

So, to find the 100th number, we must sum all the numbers from 1 to 100. To do so, we use the following formula:

`S_n=((n+1)*(n))/(2)`

And we find:

`a_(100)=S_(100)=(101*100)/(2)=5050`

The answer for 9. is 5050.