Calculate breakeven when fixed revenue being added per month v/s cost I am writing up a cost sheet for a product and I basically suck at math. Didn't know who else to turn to, so trying out Math exchange.
So, I am planning to spend $1,100 every month on advertising that will bring me $130 of additional cumulative revenue every month. Basically, it will bring me new clients that I charge $130 every month in addition to existing clients. So cost is fixed monthly, but revenue is cumulative.
How do I calculate when my cost v/s revenue breaks even, and my profit from thereon?
If I put this up on a spreadsheet, it looks something like this:

Month | Revenue | Cost
1   |       130    |      1100
2   |       260    |      1100
3   |       390    |     1100
4   |       520    |     1100
5   |       650    |     1100
6   |       780     |     1100

Total After 6 months:
Revenue: 2,730
Cost: 66,000
 A: The total cost after $n$ months is $C_{n}=1100n$ and the total revenue is $
R_{n}=130\times \frac{n(n+1)}{2}$, because
$$\begin{eqnarray*}
R_{n} &=&130\times 1+130\times 2+130\times 3+\ldots +130\times n \\
&=&130\left( 1+2+3+\ldots +n\right)  \\
&=&130\times \frac{n(n+1)}{2},
\end{eqnarray*}$$
where I used the value of the sum $$1+2+3+\ldots +n=\frac{n(n+1)}{2}.$$
Equating $C_{n}=R_{n}$
$$1100n=130\times \frac{n(n+1)}{2},$$
simplifying
$$1100=130\times \frac{n+1}{2}$$
and solving for $n$ yields $n=\frac{207}{13}\approx 15.92$. And so, the
breakeven month is $n=16$. Confirmation:
$$C_{16}=1100\times 16=17\,600,$$
$$R_{16}=130\times \frac{16(16+1)}{2}=17\,680.$$
The accumulated profit is $R_n-C_n$.
Here is a plot of $R_n$ (blue) and $C_n$ (sienna) versus $n$ (month)

A: You just need to extend your spreadsheet downwards.  
By month 9 ($\approx \frac{1100}{130}$) you should find revenues exceeding advertising costs, and after about twice as long you should find cumulative revenues exceeding cumulative advertising costs.
You can also consider other costs (such as manufacture), discounted cash flow, and the sustainability of your model at higher levels of sales.
