Originally Posted by

**pmh118** Yes the deposit is on the 1st of each month

and 20r(r^12 -1) all divided by (r -1)

OK; then your problem could be worded this way:

20 is deposited on the 1st of each month over 1 year (so Jan 1st to Dec 1st).

The annual interest rate is 12x% compounded monthly, hence x% per month.

Therefore, as at Dec 31st, the first payment has increased in value to:

20 r^12 , where r =1 + x/100.

Show that the total value, as at Dec 31st, is:

20r(r^12 -1) / (r-1).

If the rate is 1/2% per month (hence r = 1.005), calculate the toal interest

received during the 12 months.

I'll give you a shorter example (hate typing!): 100 deposited for 6 months

(Jan 1 to Jun 1) at a rate of 1% per month (hence r = 1.01):

Code:

Date Deposit Interest Balance
Jan.1 100.00 .00 100.00
Feb.1 100.00 1.00 201.00
Mar.1 100.00 2.01 303.01
Apr.1 100.00 3.03 406.04
May.1 100.00 4.06 510.10
Jun.1 100.00 5.10 615.20
Jun.30 .00 6.15 621.35
S = 100(1.01^6) + 100(1.01^5) + 100(1.01^4) + 100(1.01^3) + 100(1.01^2) + 100(1.01^1)
S(1.01) = 100(1.01^7) + 100(1.01^6) + 100(1.01^5) + 100(1.01^4) + 100(1.01^3) + 100(1.01^2)
S(1.01) - S = 100(1.01^7) - 100(1.01^1)
S(1.01 - 1) = 100(1.01^7 - 1.01^1)
S = 100(1.01)(1.01^6 - 1) / (1.01 - 1) = 621.35

Above shows you how to "explain" the formula.

As far as the total interest paid, it's simply:

Final balance - total deposits; in my example: 621.35 - 100*6 = 21.35

Hope that's enough to make your day!