Question

you deposit $2000 in an account that pays 8% interest compounded semiannually. after 2 years, the interest rate is increased to 8.32% compounded quarterly. what will be the value of the account after a total of 4 years?
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the value of the account will be $\square$
(round to the nearest dollar as needed.)

Explanation:

Step1: Calculate value after 2 years (semiannual compounding)

The compound - interest formula is $A = P(1+\frac{r}{n})^{nt}$, where $P$ is the principal amount, $r$ is the annual interest rate (in decimal), $n$ is the number of times interest is compounded per year, and $t$ is the number of years.

For the first 2 years:

• $P=\$2000$
• $r = 0.08$ (since 8% = 0.08)
• $n = 2$ (compounded semiannually)
• $t = 2$

Substitute these values into the formula:
$A_1=2000(1 +\frac{0.08}{2})^{2\times2}=2000(1 + 0.04)^{4}$

Calculate $(1 + 0.04)^{4}=1.04^{4}\approx1.16985856$

Then $A_1 = 2000\times1.16985856=\$2339.71712$

Step2: Calculate value after next 2 years (quarterly compounding)

Now, the principal for the next 2 - year period is $A_1=\$2339.71712$

For the next 2 years:

• $P = A_1=\$2339.71712$
• $r=0.0832$ (since 8.32% = 0.0832)
• $n = 4$ (compounded quarterly)
• $t = 2$

Use the compound - interest formula again: $A = P(1+\frac{r}{n})^{nt}$

Substitute the values:
$A_2=2339.71712(1+\frac{0.0832}{4})^{4\times2}=2339.71712(1 + 0.0208)^{8}$

First, calculate $1+0.0208 = 1.0208$

Then, $1.0208^{8}\approx1.0208^{8}$. We can calculate this as follows:

$1.0208^{2}=1.0208\times1.0208 = 1.04163264$

$1.0208^{4}=(1.04163264)^{2}\approx1.084960$

$1.0208^{8}=(1.084960)^{2}\approx1.177127$

Now, $A_2=2339.71712\times1.177127\approx2339.71712\times1.177127$

$2339.71712\times1.177127\approx2339.71712\times(1 + 0.177127)=2339.71712+2339.71712\times0.177127$

$2339.71712\times0.177127\approx414.44$

$2339.71712 + 414.44\approx2754.16$ (more accurately, using direct multiplication: $2339.71712\times1.177127\approx2754$ (rounded to the nearest dollar))

Answer:

$\$2754$