Equation that handles diminishing returns I've tried desperately to figure this out, but to no avail.
I need an equation that will effectively reduce a number by $0.2 \%$ each time it is added to itself (The original number).
In other words, the first number will be $100$, then the second number will be $199.8 = (100 + 99.8)$, and then the third number will be $299.4 = (100 + 99.8 + 99.6)$.
I've tried things like $Y = 100 * X - 0.2 * (X - 1)$ but that doesn't work.
Thanks in advance for any help!
 A: Let $x_0$ be the first number and $\alpha$ be the geometric ratio.
Then we have
$$x_n = \sum_{i=0}^{n} x_0 \alpha^i = x_0 \frac{\alpha^{n+1}-1}{\alpha - 1}.$$ For the example you gave we have $\alpha = 0.998$ and $x_0 = 100$.
For those of you who don't like $0$ based indexing, we have
$$y_n = y_1 \sum_{i=1}^n \alpha^{i-1} = y_1 \frac{\alpha^{n}-1}{\alpha - 1}.$$
Note that $99.8$ less $0.2\%$ is $99.6004$ not $99.6$ as stated in the question.  
A: Let your $100$ be x and $a=0.2$. Then you have the following sequences:
$b_1=(1-0a)x, b_2=(1-1a)x, b_3=(1-2a)x,\ldots , b_n=(1-(n-1)a)x$
The sum of these sequences is $n\cdot x - \sum_{i=0}^{n-1} i\cdot a=n\cdot x - a\cdot \sum_{i=0}^{n-1} i$.
$\sum_{i=0}^{n-1} i=\frac{(n-1)\cdot n}{2}$ is a well known relation.
$\color{blue}{\texttt{numerical example}}$
$a=0.2\%\cdot 100=0.2, \ x=100, \ n=3$
The formula in total is
$n\cdot  x-a\cdot \frac{(n-1)\cdot n}{2}=3\cdot 100-0.2\cdot\frac{(3-1)\cdot 3}{2}=300-0.2\cdot\frac{2\cdot 3}{2}=300-0.6=\boxed{299.4}$
