How do you write a formula for Compound Interest when you are also depositing a set amount each time it compounds?

Question

My cousin plays this game called Alien Shooter and he asked me this and I couldn't figure it out. So basically you get $10$ coins right off the bat, even if you die in the first round. After every round you beat you get an additional $10$ coins and the total value also increases by $10\%$. When I tested it I got values like this (round $0$ is dying on the first round):

Round  0  1  2  3  4  5   6   7   8   9  10  11  12  13  14  15
Coins 10 22 36 52 70 90 112 136 162 190 220 252 286 322 360 400 etc.

I thought this would be pretty simple but for some reason I can't figure out what the formula would be.

Answer 1

Despite the recurrence $x_{n+1} =\frac{11}{10}(x_{n} +10)$ is solid (it matches the description of the problem), it doesn't take into account round-ups.

That being said, the given values don't behave like an exponential, even with some rounding. The first twelve values behave just like a quadratic: $$f(n) =n^2 +11n +10 =(n+10)(n+1).$$ I'm prone to believe the OP values at $n=14$ and $n=15$ where meant to be $100$ units greater. \begin{array}{l|rr} n & 0& 1& 2& 3& 4& 5& 6& 7& 8& 9& 10& 11& 12& 13& 14& 15\\ \hline \text{OP}&10&22&36&52&70&90&112&136&162&190&220&252&284&320&258&298\\ f(n) &10&22&36&52&70&90&112&136&162&190&220&252&286&322&360&400\\ \text{diff.}&&&&&&&&&&&&&+2&+2&102&102 \end{array}

Answer 2

Take $x_{n+1}=\frac{11}{10}(x_{n}+10)$

Then $x_{n+2}=\frac{11}{10}(\frac{11}{10}(x_{n}+10))+11=(\frac{11}{10})^2x_{n}+\frac{11^2}{10}+11$

You can continue this for the general form: $$x_n=\bigg(\frac{11}{10}\bigg)^nx_0+\frac{11^{n+1}}{10^n}+\frac{11^n}{10^{n-1}}+\cdots+\frac{11^2}{10}+11$$

Since $x_0=0$, we are only interested in the other part of the series, which we can summarize as: $$x_n=\sum_{r=0}^{n}{\frac{11^{r+1}}{10^r}}$$

Answer 3

The future value of an annuity problem.

$FV = (Payment) + (Payment)(1+r) + (Payment)(1+r)^2 +\cdots+(Payment)(1+r)^n = (Payment) \frac {(1+r)^{n+1}-1}{r}$