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 | |

_________________________________________________________________________________ | ||||||||

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

What about a first payment 45 days after the loan is made? How can the amount be calculated when the payment is due at 1.5, 2.5, 3.5, etc.? It is much easier to assume the loan was made at

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