Expert: Paul Klarreich - 12/20/2006

Question
I think this fits your expertise, but please excuse me if I'm in the wrong forum. I'm not a student and am not compelled to turn in this problem to a teacher anytime soon, so please feel free to think about this and ask me questions if I don't articulate myself well enough.

I'm trying to figure out a more concise formula for figuring out how much $89 dollars invested bi-weekly over a 20 year period with 5% interest would eventually accumulate into, if left untouched.

Here's what I came up with:

(((((89)1.05) + 89)1.05 + 89)1.05 +89)1.05 + 89)1.05 + 89....{And so on for 520 times)

Yep, it's messy.  Is there more concise way of writing this calculation, aside from what I just did?  Also is there a way to write a formulate that could incorporate a variable that would allow me to calculate the sum after 10 years?  30 years?  At 3%?  6%?  

No, I didn't study math after high school at all, so this explains why this is the best I could do.  I do apppreciate any help you can offer.  Thanks!  

Answer
Questioner:  Augusto
Category:  Calculus
 
Subject:  compound interest
Question:  I think this fits your expertise, but please excuse me if I'm in the wrong forum. I'm not a student and am not compelled to turn in this problem to a teacher anytime soon, so please feel free to think about this and ask me questions if I don't articulate myself well enough.

I'm trying to figure out a more concise formula for figuring out how much $89 dollars invested bi-weekly over a 20 year period with 5% interest would eventually accumulate into, if left untouched.

Here's what I came up with:

(((((89)1.05) + 89)1.05 + 89)1.05 +89)1.05 + 89)1.05 + 89....{And so on for 520 times)

Yep, it's messy.  Is there more concise way of writing this calculation, aside from what I just did?  Also is there a way to write a formula that could incorporate a variable that would allow me to calculate the sum after 10 years?  30 years?  At 3%?  6%?  

No, I didn't study math after high school at all, so this explains why this is the best I could do.  I do apppreciate any help you can offer.  Thanks!
..........................................................
Hi, Augusto,

Indeed there is a way.  Now problems like this always have an 'off-by-one' difficulty: In this case, do you put in the last payment or not?  If you were paying off a car loan or a mortgage, I think you would, so that is what I will assume.

WARNING: THE FOLLOWING DISCUSSION MAY CONTAIN FRACTIONS AND OTHER MATERIAL INAPPROPRIATE FOR CERTAIN COMPUTING SYSTEMS.  VIEW IT IN A FIXED-SIZE FONT, SUCH AS COURIER.


So the problem becomes:

You make  n payments each year.  (in this case n = 26)
Each payment is  $P.             (in this case P = 89)
The interest rate is  r.         (here  r = 0.05)
The time period is y.            (here  y = 20)

Suppose some payment is made during the time period.

The 'compounding interval' [C.I.] is  1/n years, (that's your two weeks), and in that interval the interest is paid at  r/n percent.  [here that's 0.05/26]

The total number of C.I.'sin the example is the number of years times the number of payments per year:

There are  n*y total intervals.  

How many 'bumps' will this one payment receive?  Assume this is payment number k, where k is a number between 0 and n*y.

[We denote the first payment as number 0 and the last as number ny.]

So what is the growth of this payment?  The interest is compounded  ny - k  times, so this one payment grows to

P(1 + r/n)^(ny - k)  dollars.

So you have a summation like this:

The first payment (payment zero) grows to:

P(1 + r/n)^(ny)  dollars.

The last payment (payment ny) grows to:

P(1 + r/n)^(0)  dollars. [doesn't grow at all]

The sum is:

P[(1 + r/n)^0 + (...)^1 + (...)^2 + ... + (1 + r/n)^(ny)]

Aha! That's a Geometric Series, and there is a formula: (you could look it up, said the old guy, but it's more fun to derive it yourself.)

     S = 1 + a + a^2 + ... + a^n
    aS =     a + a^2 + ... + a^n + a^n+1
--------------------------------------- Subtract
aS - S  = - 1 + a^n+1
   a^n+1 - 1
S = ----------  [standard formula for sum of a G.S.]
     a - 1                                
Back to the example:

The sum is:  (again)

P[(1 + r/n)^0 + (...)^1 + (...)^2 + ... + (1 + r/n)^(ny)]
    (1 + r/n)^ny+1  -  1
= P[ --------------------- ]
       1 + r/n - 1

    (1 + r/n)^ny+1  -  1
= P[ --------------------- ]
        r/n  

And that's basically it.  Now just take your calculator on that, put in your values:

r = 0.05  (interest rate)
n = 26    (pmts / year)
y = 20    (years)
P = $89   (payment)

or whatever other values you choose.