scenario #1: raul is a saver. he sets aside $100 per month during his career of 40 years to prepare for…

scenario #1: raul is a saver. he sets aside $100 per month during his career of 40 years to prepare for retirement. he does not like the idea of investing because he prefers to minimize his risk as much as possible, so he puts his money in a savings account which earns 1.5% interest per year. using the compound interest calculator, what is the total balance in the account after 40 years? *round your answer to the nearest dollar and input numbers only (no symbols).

scenario #1: raul is a saver. he sets aside $100 per month during his career of 40 years to prepare for retirement. he does not like the idea of investing because he prefers to minimize his risk as much as possible, so he puts his money in a savings account which earns 1.5% interest per year. using the compound interest calculator, what is the total balance in the account after 40 years? *round your answer to the nearest dollar and input numbers only (no symbols).

Answer

Explanation:

Step1: Convert years to months

$40\times12 = 480$ months

Step2: Convert annual - interest rate to monthly - interest rate

The annual interest rate $r = 1.5%=0.015$, so the monthly interest rate $i=\frac{0.015}{12}=0.00125$

Step3: Use the future - value of an ordinary annuity formula

The formula for the future - value of an ordinary annuity is $F = A\times\frac{(1 + i)^{n}-1}{i}$, where $A = 100$ (monthly payment), $i = 0.00125$ (monthly interest rate), and $n = 480$ (number of periods)

$F=100\times\frac{(1 + 0.00125)^{480}-1}{0.00125}$

First, calculate $(1 + 0.00125)^{480}$: Let $x=(1 + 0.00125)^{480}$, using the formula $a^{b}=e^{b\ln(a)}$, we have $\ln(x)=480\times\ln(1.00125)$ $\ln(1.00125)\approx0.001249$, so $\ln(x)=480\times0.001249 = 0.59952$ $x = e^{0.59952}\approx1.821$

Then, $F = 100\times\frac{1.821 - 1}{0.00125}=100\times\frac{0.821}{0.00125}=100\times656.8 = 65680$

Answer:

65680