# Invest a dollar at 6% interest compounded monthly

1. Apr 1, 2009

### reb659

1. The problem statement, all variables and given/known data

If you invest a dollar at 6% interest compounded monthly, it amounts to (1.005)^n, where n=# of months. If you invest 10$at the beginning of each month for 10 years (120 months), how much do you have at the end of the 10 years? 2. Relevant equations sum of a geometric series - a/(1-r) 3. The attempt at a solution The only examples of this I've seen were when the same amount is being compounded. I don't understand how I would sum up the total if money is being added each month. Last edited: Apr 1, 2009 2. Apr 1, 2009 ### lanedance hi reb659 i'd try writing out the calc for value of money at each month, for the first few months, and see if you can pick the pattern - 3. Apr 1, 2009 ### reb659 but how does the "If you invest a dollar at 6% interest compounded monthly, it amounts to (1.005)^n, where n=# of months" help me solve it? 4. Apr 1, 2009 ### meiso I understand your confusion. You are correct that this problem is different from compound interest problems we normally see. If we had invested$10 from the beginning, and left it alone, after 120 months we'd have 10*(1.005)^120 = $18. If we had invested all$1200 ($10 * 120 months) from the beginning, we'd have 1200*(1.005)^120 =$2183.28.

When we add $10 at the beginning of each month, however, we cannot use the simple formula above. In your solution to this problem, though, you will still be using the factor 1.005 to calculate the interest on the money. It helps to think of the interest on each$10 deposit as being calculated separately. So, the first $10 earns interest for the first 120 months, the second ten dollars earns interest for 119 months, etc. When you come up with the expression for this calculation, you will begin to see why this problem pertains to geometric series. I would take the above advice of lancedance and write out the first, say, four or five months, using variables for 10 and 1.005 to simplify the algebra. It is often easier to spot a pattern when too many numbers aren't involved. So, for example, if we choose P=10 and R=1.005, the total money on the first$10 deposit will be PR^120, and on the next PR^119.

One important note: The formula you have listed in your original posting is for the sum of an infinite geometric series. Since they are asking you for a sum of money after 120 months, you are dealing with a finite series. The sum of the first n terms of a finite geometric series is equal to:

$$\frac{a(1-r^{n})}{1-r}$$

where r is the common ratio of the series and a is the first term of the series.

Last edited: Apr 1, 2009
5. Apr 1, 2009

### lanedance

this gives you the monthly rate as 1.005 or 0.5% increase, every month

this comes from 6% compounded monthly gives, 6% divided by 12 months gives 0.5% per month

6. Apr 2, 2009

### reb659

Thanks a ton for the help guys. I was looking way too hard to find a general formula for this situation.

I came with:

∑10*1.005^n from n=1 to n=120
= 10(1.005(1-1.005^120)/(1-1.005))
= 1646.987

It seems reasonable, since it should be less than the total sum if all 1200$were invested at once in the beginning. Last edited: Apr 2, 2009 7. Apr 2, 2009 ### meiso Yea, that's what the problem was going for. I got a slightly different number though:$1628.79. The only thing you have to remember is since the first term of a geometric series is not supposed to have the ratio in it, we have to factor 10(1.005) out of the series if you want to use that formula for the sum, which it looks like you did. However, when you do that, you must remember that factoring out the 1.005 reduces the highest power of the series by one (120->119).

So, correct me if I'm wrong, but I believe it would be

10(1.005(1-1.005^119)/(1-1.005))

which gives the slightly different answer I got.

8. Apr 2, 2009

### reb659

I looked up the formula for sum of a finite geometric series which is

a + ar + ar^2 + ....ar^(n-1) = a(1-r^n)/(1-r)

so 10+10(1.005) + 10(1.005)^2 + .... 10(1.005)^120 = 10(1-1.005^121)/(1-1.005)

I don't need the first term as part of the sum so I subtract 10 from both sides to get:

10(1.005) + 10(1.005)^2 + .... 10(1.005)^120 = 10(1-1.005^121)/(1-1.005)-10

= 1646.976

I don't understand why you factored out the first term. Why is that necessary?

Last edited: Apr 2, 2009
9. Apr 2, 2009

### meiso

Ok. I see what you did now by subtracting the ten. My mistake!

Last edited: Apr 2, 2009
10. Apr 2, 2009

### reb659

Phew! I was scared I had gotten it wrong again for a second. Thanks for all your help!

11. Apr 7, 2009

### reb659

Gah! I actually got the answer from the professor today and it turns out the answer was 1638.79. I could have sworn I did it right....

Anyone see the mistake?

12. Apr 7, 2009

### meiso

I believe your professor is actually wrong on this one.

Let R = 1.005
The answer he/she gave you is actually the value that results from $$\frac{10(1-R^{120)}}{1-R}$$. This is equal to 1638.79 rounded to two decimal places. But this expression would correspond to the sum of the geometric series:

10 + 10R^1 + 10R^2 + ... + 10R^118 + 10R^119.

This is clearly wrong, since we are missing the ten dollars that accumulates interest for 120 months, and we have a ten dollar amount (at the beginning of the series) that never accumulates any interest. This situation does not match up with the premises given in the problem statement, where themanner in which it is written leads us to believe that the interest is accrued at the end of each month.

I would bring this up to the professor, show him/her your method of solution, and ask that he/she proves you wrong.

I am fairly sure that we have thought about this correctly, but I would love to see how we are wrong if we are.

13. Apr 9, 2009

### reb659

I talked with my professor and he verified that his solution was wrong and ours was correct. Thanks again for your help!