Some guy opens a bank account with an initial amount of $\$1,000$. Each month he deposits $\$200$ and the bank gives him a monthly interest of $6\%$. I want to find the closed formula.
Given this, we can derive the recurrence-relation: $y_n=1.06y_{n-1} + 200, y_0 = 1000, n \ge 1$. Now, I know that it can also be written as: $y_n=1.06^ny_0 + 200n = 1.06^n*1000 + 200n$. Or so I thought. When I try this formula for $n=2\ or\ n=3$, the formula is not equivalent to the recurrence-relation.
I think I know this has something to do with the fact that the coefficient $1.06$ only applies to the initial cash, and not the deposits he made after that, and so I tried the formula: $$y_n = 1.06^n*1000 + 1.06^{n-1}*200n$$ However, this formula is also wrong, maybe even more so. I'm in need of some help to get past this obstacle.
Own thoughts on how to progress (in a structural manner)
$y_n = 1.06y_{n-1}+200, y_0 = 1000, n \ge 1$
$y_1 = 1.06y_{0} + 200 = 1.06*1000 + 200$
$y_2 = 1.06y_{1} + 200 = 1.06(1.06*1000 + 200) + 200 = 1.06^2*1000 + 1.06*200 + 200$
$\begin{align} y_3 = 1.06y_{2} + 200 &= 1.06(1.06^2*1000+1.06*200+200) + 200\\ &=1.06^3*1000 + 1.06^2*200 + 1.06*200 + 200 \\ &= 1.06^3*1000+1.06^2*200+1.06*200+1.06^0*200 \end{align}$
Don't know if this even helps me. Any help is appreciated, many thanks.
let:
$M = \mbox{monthly deposit}$
$R = \mbox{interest rate}$
$S = \mbox{starting capital}$
the recurrence-relation is:
$$ \mbox{given: } n \in \mathbb{N}\\ \begin{cases} y_0 = S& \mbox{if } n=0\\ y_n = Ry_{n-1}+M &\mbox{otherwise} \end{cases} $$ by calculating manually the first few terms, and performing some algebra one can find that:
$$ y_n=S \cdot R^n + M \cdot \sum_{i=0}^{n-1} R^i $$ now let's remember the Geometric Series: $$ \sum_{i=0}^{n} x^i =\frac{1-x^{n+1}}{1-x} $$ we will use this to explicitly calculate $\sum_{i=0}^{n-1} R^i$, substituting and correcting the upper bound gives: $$ y_n=S \cdot R^n + M \cdot \sum_{i=0}^{n-1} R^i \iff \boxed{y_n=S \cdot R^n + M \cdot \frac{1-R^{n}}{1-R}} $$