question 2\nsuppose $1,250 is deposited into an account each quarter. the account grows at an annual rate of…

Explanation:

Step1: Identify the relevant formula

The formula for the future - value of an ordinary annuity is $F = A\times\frac{(1 + r)^{n}-1}{r}$, where $A$ is the amount of each payment, $r$ is the interest rate per period, and $n$ is the number of periods. The annual interest rate $i = 10.0%=0.10$. Since it is compounded quarterly, the interest rate per quarter $r=\frac{0.10}{4}=0.025$. The number of years $t = 45$ years. The number of quarters $n = 45\times4=180$ quarters. The amount of each quarterly payment $A = 1250$.

Step2: Substitute the values into the formula

$F=1250\times\frac{(1 + 0.025)^{180}-1}{0.025}$. First, calculate $(1 + 0.025)^{180}$. Let $x=(1 + 0.025)^{180}$. Using the formula $a^{b}=e^{b\ln(a)}$, we have $\ln(x)=180\times\ln(1.025)$. $\ln(1.025)\approx0.0247$, so $\ln(x)=180\times0.0247 = 4.446$. Then $x = e^{4.446}\approx85.947$. $(1 + 0.025)^{180}-1\approx85.947-1 = 84.947$. $\frac{(1 + 0.025)^{180}-1}{0.025}=\frac{84.947}{0.025}=3397.88$. $F = 1250\times3397.88=4247350$.

Answer:

4247350