How to calculate the sum of a series, whose next value is dependent on the previous value? Consider the below scenario
A -> Amount ; C->Constant; V1 -> Value of first term ; V2 -> Value of 2nd term and so on up to n terms
Now, I need to get the sum of the below series up to n terms. For e.g.
A*C + (A-V1)*C + (A-V2)*C + (A-V3)*C + .... n terms
Where, V1 = A*C ; V2 = (A-V1)*C ; V3 = (A-V2)*C and so on ...
I tried to do it manually on paper, but am getting huge confusing equations. I am sure there must be an easy way around.
 A: Noting that:
$$
\begin{aligned} 
& \frac{V_0}{A\cdot C} = \frac{0}{A \cdot C} = 0 \\
& \frac{V_1}{A\cdot C} = \frac{\left(A - V_0\right)\cdot C}{A \cdot C} = 1 \\
& \frac{V_2}{A\cdot C} = \frac{\left(A - V_1\right)\cdot C}{A \cdot C} = 1 - C \\
& \frac{V_3}{A\cdot C} = \frac{\left(A - V_2\right)\cdot C}{A \cdot C} = 1 - C + C^2\\
& \dots \\
& \frac{V_m}{A\cdot C} = \frac{\left(A - V_{m-1}\right)\cdot C}{A \cdot C} = 1 - C + C^2 + \dots + C^{m-1} \\
\end{aligned}
$$ multiplying both members by $1 + C$, we get:
$$
\begin{aligned} 
& \frac{1 + C}{A\cdot C}\,V_0 = 0 \\
& \frac{1 + C}{A\cdot C}\,V_1 = 1 + C \\
& \frac{1 + C}{A\cdot C}\,V_2 = 1 - C^2 \\
& \frac{1 + C}{A\cdot C}\,V_3 = 1 + C^3\\
& \dots \\
& \frac{1 + C}{A\cdot C}\,V_m = 1 - (-C)^m \\
\end{aligned}
$$
ie:
$$V_m = \frac{A \cdot C}{1 + C} \cdot \left(1 - (-C)^m\right).$$
This done, all that remains is to calculate the following summation:
$$\sum_{m = 0}^n V_m = \frac{A \cdot C}{1 + C} \cdot \left(\sum_{m = 0}^n 1 - \sum_{m = 0}^n (-C)^m\right)$$
ie:
$$S_n = \frac{A \cdot C}{1 + C} \cdot \left(1 + n - \frac{1 - (-C)^{1 + n}}{1 + C}\right),$$ 
which is what is desired.
In particular, fixed $A = 700\,000$ and $C = 0.015$, tabulating $n$ and $S_n$ we obtain:

If the formula of the partial sums of the geometric progression is not remembered:
$${S^*}_n = 1 + q^1 + q^2 + q^3 + \dots + q^n$$
then multiplying both members by $q$, we get:
$$q \cdot {S^*}_n = q^1 + q^2 + q^3 + q^4 + \dots + q^{1+n}$$
and subtracting the two equations member to member, we have:
$$(1 - q) \cdot {S^*}_n = 1 - q^{1+n}$$
ie:
$${S^*}_n \equiv \sum_{m = 0}^n q^m = \frac{1 - q^{1 + n}}{1 - q}\,,$$
as we wanted to prove.
A: $$V_{n+1}=C(A-V_n)=CA-CV_n$$ with $V_0=0$ is a so-called arithmetico-geometric progression.
There is a standard formula for such series. Lookup Arithmetico–geometric_sequence#Sum_of_the_terms on Wikipedia.
