How Does Compound Interest Affect Birthday Savings Over 20 Years?

[Johnny Leong]

Question:
A man puts $10 in the bank for his son on each of his birthdays from the first to the twentieth inclusive. If the money accumlates at 3% compound interest, what is the toatl value on the son's twenty-first birthday?

My answer is like this:
a = 10, r = 1.03, n = 20
Total value = a * (r^n - 1) / (r - 1) = 10 * (1.03^20 - 1) / 0.03 = 269 (approx.)

But the answer is 276, what is the problem?
The $10 of the first year will immediately be counted for interest or not? And also how to define the final year, it is the 20th year or the 21st year?

 

[Janitor]

Hmmmmm

I just did it in a spreadsheet, and I got $286.76, which doesn't match either answer.

I used Microsoft Works, so I am prepared to blame Bill Gates if my answer is wrong.

Aha! Maybe he doesn't put $10 in on the 21st birthday, so subtract $10.00 from $286.76, and you get $276.76. Yeah, I think that's what is going on. That last year the account grows, but no new principle is put into the account.

 

[HallsofIvy]

Janitor, the original problem said "A man puts $10 in the bank for his son on each of his birthdays from the first to the twentieth inclusive" so there is no "maybe" about it.

Johnny Leong, you have the formula wrong. The sum of the geometric series \Sigma_{i=0}^n{ar^i} is a\frac{1-r^{n+1}}{1-r}, not a\frac{1-r^n}{1-r}.

 

[Chen]

Is it just me or are you all missing something?

If the "man puts $10 in the bank for his son on each of his birthdays", then the series should be defined like this:

a_0 = 0
a_{n+1} = a_n 1.03 + 10

And if that is the case, the series is no longer purely geometric so the formula above cannot be used to calculate the sum.

 

[Johnny Leong]

Chen, that means on the son's first birthday. He has $10 and on his 21st birthday, he will be having a21 = 1.03 * a20, don't have another $10 added.
The expanded form of a21 = 1.03 * (10 + 1.03 * 10 + 1.03^2 * 10 + ... + 1.03^19 * 10) = 10 * (1.03^21 - 1) / (1.03 - 1) - 10 = 276.