Suppose that you earned a bachelor's degree and now you're teaching high school. The school district offers teachers the opportunity to take a year off to earn a master's degree. To achieve this goal, you deposit $$ \$ 3000 $$ at the end of each year in an annuity that pays $$ 4.5 \% $$ compounded annually. a. How much will you have saved at the end of five years? b. Find the interest.

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To find the total amount saved at the end of five years and the interest earned, we can use the formula for the future value of an annuity:

$$ FV = P \times \left( \frac{(1 + r)^n - 1}{r} \right) $$

where:
- $$ FV $$ is the future value of the annuity,
- $$ P $$ is the annual deposit amount,
- $$ r $$ is the annual interest rate,
- $$ n $$ is the number of years.

Given:
- $$ P = \$3000 $$ (annual deposit amount),
- $$ r = 4.5\% = 0.045 $$ (annual interest rate),
- $$ n = 5 $$ (number of years).

a. To find the total amount saved at the end of five years:
$$ FV = 3000 \times \left( \frac{(1 + 0.045)^5 - 1}{0.045} \right) $$

b. To find the interest earned:
$$ Interest = FV - P \times n $$

Let's calculate the total amount saved at the end of five years and the interest earned.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{3000\left(\left(1+0.045\right)^{5}-1\right)}{0.045}$$
- step1: Add the numbers:
  $$\frac{3000\left(1.045^{5}-1\right)}{0.045}$$
- step2: Convert the expressions:
  $$\frac{3000\left(\left(\frac{209}{200}\right)^{5}-1\right)}{0.045}$$
- step3: Subtract the numbers:
  $$\frac{3000\times \frac{209^{5}-200^{5}}{200^{5}}}{0.045}$$
- step4: Multiply the numbers:
  $$\frac{\frac{3\times 209^{5}-3\times 200^{5}}{4096\times 5^{7}}}{0.045}$$
- step5: Convert the expressions:
  $$\frac{\frac{3\times 209^{5}-3\times 200^{5}}{4096\times 5^{7}}}{\frac{9}{200}}$$
- step6: Multiply by the reciprocal:
  $$\frac{3\times 209^{5}-3\times 200^{5}}{4096\times 5^{7}}\times \frac{200}{9}$$
- step7: Rewrite the expression:
  $$\frac{3\left(209^{5}-200^{5}\right)}{4096\times 5^{7}}\times \frac{200}{9}$$
- step8: Reduce the numbers:
  $$\frac{209^{5}-200^{5}}{512\times 5^{7}}\times \frac{25}{3}$$
- step9: Rewrite the expression:
  $$\frac{209^{5}-200^{5}}{512\times 5^{7}}\times \frac{5^{2}}{3}$$
- step10: Reduce the numbers:
  $$\frac{209^{5}-200^{5}}{512\times 5^{5}}\times \frac{1}{3}$$
- step11: Multiply the fractions:
  $$\frac{209^{5}-200^{5}}{1600000\times 3}$$
- step12: Multiply:
  $$\frac{209^{5}-200^{5}}{4800000}$$
The total amount saved at the end of five years is approximately $16412.13.

Now, let's calculate the interest earned.
Calculate the value by following steps:
- step0: Calculate:
  $$16412.13-3000\times 5$$
- step1: Multiply the numbers:
  $$16412.13-15000$$
- step2: Subtract the numbers:
  $$1412.13$$
The interest earned is approximately $1412.13.

Therefore, at the end of five years, you will have saved a total of $16412.13, and the interest earned will be $1412.13.
At the end of five years, you will have saved a total of \$16,412.13, and the interest earned will be \$1,412.13.

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