ok i know the compound intrest formula is p=n*r^n but what is the formula if you add every month?

Answer 1

p=(i)(1+(r/12))^12n
i is initial deposit
r is interest rate
n is # of years

Answer 2

I'll presume that n is the number of compounding periods. So if the time the principle is left in the account is 2 years, then n=(12)(2)=24

The variable "r" is the annual percentage rate divided by the number of compounding periods in a year, Say you APR was 6%. The "r" you would use in the formula is .06/12 or .005 or 0.5%.

Answer 3

Use a LaTeX editor to see the equations clearly. Use the simple online form to see the equations displayed:

http://www.tlhiv.org/cgi-bin/LaTeXpreviewer/index.cgi

\[
A = P(1+r)^n
\]

is the compound interest formula.

If however you add every month p amount:

\[
A = \sum_{k=1}^{n} k p {(1+r)}^k
\]


In simpler terms:
\[
A = p(1+r) + 2p(1+r)^2 + 3p(1+r)^3 + ...
\]
This is an A.G.P

The solution for this would thus be:
\[
A = p \left\{ \frac{(1+r) (1-(1+r)^n)}{r^2} + \frac{(1+r)^{n+1}}{r} \right\}
\]

Here p is the monthly installment and r is the rate of interest.

To understand how this is actually coded, use a LaTeX previewer link provided above

Answer 4

The actual formula is P(1+r)^n. If you add every month, then n is the number of months and r is the amount added every month. Essentially you just replace years with months.

Answer 5

Let p = principal, r = annual interest rate, n = no. of years, i = accumulated interest for the period.

Answer: determine the accumulated interest for every period:
i = (p * [{1 + |r / 12|} * {1 + |r / 12|}]^[12n - 1]) - p