The formula for finding the periodic payments for a loan is easy to find.
Let Pn be the amount owed at period n. At time period 0, we are given an initial principal amount P. After one time period, our balance increases by an amount equal to interest, i, times P, but our balance is also decreased by the amount of the periodic payment, M. So, after one period, we have P1 = P0(1+i) - M. The period 2 balance is computed from the period 1 balance similarly as given in the table below.
| Period 0 | Period 1 | Period 2 | Period n |
| P0 = P | P1 = P0(1+i) - M | P2 = P1(1+i) - M | Pn = Pn-1(1+i) - M |
| P0 = P | P1 = P(1+i) - M | P2 = P(1+i)2 - (1+i)M - M | Pn = P(1+i)n - (1+i)n-1M . . . - (1+i)2M - (1+i)M - M |
Substituting the period n formulas into the n-1 formulas, the second row of the table gives the n period formula in terms of only the initial principal balance giving us the following formula.
| PN = P(1+i)N - M |
__ N - 1 > (1+i) n n = 0 |
This is an arithmetic series which is given by:
|
__ N - 1 > a n n = 0 |
= |
1-aN --- 1-a |
where a = (1+i). Substituting gives the result below. Lets also assume that the Nth payment is the last payment which will bring the balance to 0. Thus PN = 0.
| PN = P(1+i)N - M |
1 - (1+i)N ------ = 0 1 - (1+i) |
Solving for M gives:
| M = P(1+i)N * |
1 - (1+i) ------ 1 - (1+i)N |
= P |
(1+i)N - (1+i)N(1+i) ------------ 1-(1+i)N |
Multiplying the top and bottom of the right hand side by (1+i)-N aka multiplying by 1, gives
| M = P |
1 - (1+i) ------ (1+i)-N-1 |
= |
P i ----- 1-(1+i)-N |
which is the standard formula for loan payments.
If the periodic payments are not per year, the interest and period must be adjusted. For example, for monthly payments, if the loan is 30 year at 6% interest per year, N = 30 years * 12 months/year = 360 months and i = .06 / 12 months/year = .005 / month. The (rounded) monthly payment would be $6.00 per $1000 borrowed. For $50,000 multiply this payment amount by 50 to get the monthly payment or use the formula with P = 50,000.
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