Let's assume 10% interest every period. After the first period the account will have $1100. When interest is credited at the end of the second period, it will be not on just the original $1000 but also on the interest already earned. Instead of having a value of $1200, it will be worth $1100 * 1.1 or $1210. Likewise, it's value after three periods is $1000 * (1.10) * (1.10) * (1.10) = $1331. The number

If we use

Assume a first payment is made of amount

1) | F = | A * (1 + i)^{n - 1} | +A * (1 + i)^{n - 2} | + . . . . | +A * (1 + i)^{2} | +A * (1 + i)^{1} | +A | |

This can be calculated by multiplying both sides by (1 + i) then subtracting the first formula from it. | ||||||||

Only the end of the first formula and beginning of the new one will remain. | ||||||||

2) | F * (1+i) = | A * (1 + i)^{n} | +A * (1 + i)^{n - 1} | +A * (1 + i)^{n - 2} | + . . . . | +A * (1 + i)^{2} | +A * (1 + i)^{1} | |

1) | F = | A * (1 + i)^{n - 1} | +A * (1 + i)^{n - 2} | + . . . . | +A * (1 + i)^{2} | +A * (1 + i)^{1} | +A | |

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F * i = | A * (1 + i)^{n} | - A | ||||||

Combining the A's on the right side: | ||||||||

F * i = | A * [(1 + i)^{n}
- 1] | |||||||

Solving for A: | ||||||||

A = | F * i
| |||||||

[(1 + i)^{n} - 1] | ||||||||

Since F = P * (1 + i)^{n}, we can substitute it and A is terms of P, the value of the loan, | ||||||||

the interest rate, and n, number of periods: | ||||||||

A = | P * i * (1 + i)
^{n} | |||||||

(1 + i)^{n} - 1 | ||||||||

The top and bottom of that can be divided by (1 + i)^{n} to simplify it further: | ||||||||

A = | P * i | |||||||

1 - 1 / (1 + i)^{n} |

Interest is usually expressed as an