# Recursive Solution to Interest with Monthly Deposits

I open an account at a bank with 1% interest compounded monthly. I'm adding $100 to it at the beginning of each month (starting with month 1). (a) Set up a recurrence relation for the amount in the account at the end of n months. (b) Find an explicit formula for the amount in the account at the end of n months. (c) On what date will you be able to withdraw$10000?

(a) I was able to get the recurrence relation (I think), and set it up as

$a_n=(a_{n-1}+100)(1+\frac1{1200})$

where $a_0=0$

(b) I came up with an equation that I thought would work, but it doesn't. I created a program in C++ to test it, and I came up with ~1% error after 25 months

$(100n)(1+\frac1{1200})^n$

I was wondering if I could get some help on this one or a link to another thread or page that explains how I could solve this.

EDIT

Thanks to Gerry Myerson, I was able to find the result using $$a_n=100*\sum\limits_{i=0}^n (1+\frac1{12000})^n - 100$$

This turned into $$a_n=100(\frac{(1+\frac1{1200})^{n+1}-1}{(1+\frac1{1200})-1}-1)$$

Simplified to $$a_n=100(1200[1+\frac1{1200}]^{n+1}-1201)$$

Translated to C++ this appears to work (regardless of how hard it is to read).

(c) Finished the first two, but I'm still working on this. Assistance will still be appreciated.

Yes, this is a homework question. I'm not looking for just answers, I'm looking for useful information that will help me solve these problems in the future without any help. I have no idea how to go beyond the recursive solution.

Also, this is my first time using this site, so any help with the formatting to make it easier for other people to read would also be nice.

• Let me write $r$ for $1+(1/1200)$. So your recurrence is $a_n=(a_{n-1}+100)r$ (note: you wrote $a_0$, but I think you meant $a_n$). So, let's calculate: $a_0=0$, $a_1=100r$, $a_2=100r^2+100r$, $a_3=100r^3+100r^2+100r$, and so on. I expect you can see the pattern, write down a formula for it, and prove it works by induction. You may also note that you're getting a geometric progression, which lets you simplify the formula. Commented Sep 26, 2014 at 7:11
• @GerryMyerson Thanks for the help with part two. I'm now working on part 3. Commented Sep 26, 2014 at 9:33
• So, you take the expression you got for $a_n$, set it equal to 10000, and solve for $n$. At some point along the way, you'll want to take the logarithm of both sides, to get that $n+1$ down from the exponent. Commented Sep 26, 2014 at 10:37
• That seems incredibly painful to do. I only got a little bit into it before I realized how difficult it would be for me. I don't know how to do that log at all. Commented Sep 26, 2014 at 10:42
• Painful? You divide both sides by 100. You add 1201 to both sides. You divide both sides by 1200. Doesn't seem too painful yet. Now you get where you have to take logarithms on both sides of the equation. If you haven't studied logarithms at all, I don't see how you're meant to do the problem. Commented Sep 26, 2014 at 10:46

Let's finish this off by doing part (c). We need to find the least $n$ such that $$100(1200(1201/1200)^{n+1}-1201)\ge10000$$ Divide by 100: $$1200(1201/1200)^{n+1}-1201\ge100$$ Add 1201 to both sides: $$1200(1201/1200)^{n+1}\ge1301$$ Divide by 1200: $$(1201/1200)^{n+1}\ge1301/1200$$ Take logarithms (to whatever base you like): $$(n+1)\log(1201/1200)\ge\log(1301/1200)$$ Do another division: $$n+1\ge{\log(1301/1200)\over\log(1201/1200)}$$ Subtract 1 from both sides: $$n\ge{\log(1301/1200)\over\log(1201/1200)}-1$$ Now let your calculator have a go at those numbers on the right side: I get $$n\ge96.01$$ So, it seems you're almost there after 96 months, but you have to wait for the 97th month to get over \$10,000. [After 96 months, you have \$9,998.44, according to my calculations]