Lizette decides that starting in January she will deposit $50 into her bank account at the start of each month. Her account earns 0.25% interest per month. Interest is calculated at the end of each month. Truncate answers to two decimal places.

(a) In the middle of February, how much money is in Lizette's account?

(b) In the middle of March, how much money is in Lizette's account?

(c) In the middle of the *n*th month (where January is the 1st month, February is the 2nd month, etc.), how much money is in Lizette's account? Give your answer in closed form.

(d) In the middle of December, how much money is in Lizette's account?

Answer

This exercise is very similar to the previous one. In January, Lizette deposits $50 and earns 0.25% interest. After the interest calculation, she has

1.0025(50) = 50.125

which we truncate to $50.12.

Remember that to calculate interest we multiply by (1 + ( interest rate)) since we have to account for the initial amount of money as well as the interest earned.

(a) At the beginning of February, Lizette adds $50 to the $50.12 she had from January:

Since interest isn't calculated until the end of the month, in the middle of February she still has

50 + 50.12 = 100.12.

(b) Extend the timeline one more month:

Lizette earns interest on her $100.12, then deposits another $50. In the middle of March she has

50 + 1.0025(100.12) = 150.37.

(c) Let *M*_{n} be the amount of money in Lizette's account in the middle of the *n*th month. Then *M*_{n – 1} was the amount of money in her account in the middle of the (*n* – 1)st month.

To find *M*_{n} we calculate interest on *M*_{n – 1}, then add another $50. So

*M*_{n} = 50 + 1.0025*M*_{n – 1}.

Here are the values of *M*_{n} for some small *n*:

Since we're so good at recognizing these patterns by now, we see that *M*_{n} is the sum of a finite geometric series with *a* = 50,

*r* = 1.0025, and *n* terms:

(d) There are two ways to do this problem. We could use the formula from (c) to find *M*_{12}, or we could calculate *M*_{n} for *n* = 1, *n* = 2, all the way up to *n* = 12. Since *n* = 12 is pretty small and we want to make sure we get the right answer, we'll do it both ways.

Using the formula from (c):

All we have to do is plug in *n* = 12 to the formula we found in (c). We get

which, since we're truncating our answers, works out to $608.31 in the bank.

Using direct calculation:

We found that, in March, Lizette had

*M*_{3} = 150.37.

Now we apply the formula

*M*_{n} = 50 + 1.0025*M*_{n – 1}

nine times:

*M*_{4} = 200.74

*M*_{5} = 251.24

*M*_{6} = 301.86

*M*_{7} = 352.61

*M*_{8} = 403.49

*M*_{9} = 454.49

*M*_{10} = 505.62

*M*_{11} = 556.88

*M*_{12} = 608.27

We're 4 cents off from the number we got using the formula, because here we truncated each value of *M*_{n} along the way. The answers are close enough for us to be confident that the formula is correct.