Question

suppose you could make a single \lump sum\ deposit of $3976, in an investment that provides an annual percentage rate(apr) of 2% compounded quarterly. determine the future value(fv) of the investment after 34 years.

Explanation:

Step1: Recall compound interest formula

The formula for compound interest when compounded $n$ times a year is $FV = P(1 + \frac{r}{n})^{nt}$, where $P$ is the principal amount, $r$ is the annual interest rate (in decimal), $n$ is the number of times compounded per year, and $t$ is the number of years.
Here, $P = 3976$, $r = 0.02$ (since 2% = 0.02), $n = 4$ (compounded quarterly), and $t = 34$.

Step2: Calculate the exponent $nt$

$nt = 4\times34 = 136$

Step3: Calculate the rate per period $\frac{r}{n}$

$\frac{r}{n}=\frac{0.02}{4}=0.005$

Step4: Calculate $(1 + \frac{r}{n})^{nt}$

$(1 + 0.005)^{136}\approx(1.005)^{136}$. Using a calculator, $(1.005)^{136}\approx1.9674$ (you can calculate this using the formula for compound interest or a calculator function for exponents).

Step5: Calculate the Future Value $FV$

$FV = 3976\times1.9674\approx3976\times1.9674\approx7822.38$ (more accurately, using a calculator for the exponent: $(1.005)^{136}=e^{136\ln(1.005)}\approx e^{136\times0.0049875}\approx e^{0.6783}\approx1.9674$. Then $3976\times1.9674 = 3976\times(2 - 0.0326)=7952 - 3976\times0.0326 = 7952 - 129.62 = 7822.38$)

Answer:

The Future Value of the investment after 34 years is approximately $\$7822.38$ (more precise calculation: using a calculator for $(1.005)^{136}$:

First, calculate $\ln(1.005)\approx0.00498752$

Then $136\times0.00498752 = 0.678302$

Then $e^{0.678302}\approx1.96743$

Then $3976\times1.96743 = 3976\times1.96743$. Let's calculate $4000\times1.96743 = 7869.72$, subtract $24\times1.96743 = 47.21832$, so $7869.72 - 47.21832 = 7822.50168\approx\$7822.50$)

So the more accurate answer is approximately $\$7822.50$ (depending on the calculator's precision, it can vary slightly, but around $\$7822$ to $\$7823$).