$100 are deposited into an account with a 6.5% interest rate, compounded weekly (52 times per year).\nfind…

$100 are deposited into an account with a 6.5% interest rate, compounded weekly (52 times per year).\nfind the accumulated amount after 40 years.\nhint: $a = p(1+\frac{r}{k})^{kt}$\nround your answer to the nearest cent (hundredth).

$100 are deposited into an account with a 6.5% interest rate, compounded weekly (52 times per year).\nfind the accumulated amount after 40 years.\nhint: $a = p(1+\frac{r}{k})^{kt}$\nround your answer to the nearest cent (hundredth).

Answer

Explanation:

Step1: Identify the values of P, r, k, and t

$P = 100$ (principal amount), $r=0.065$ (annual interest rate as a decimal), $k = 52$ (number of compounding periods per year), $t = 40$ (number of years)

Step2: Substitute values into the compound - interest formula

$A=P(1 +\frac{r}{k})^{kt}=100(1+\frac{0.065}{52})^{52\times40}$

Step3: First, calculate the value inside the parentheses

$\frac{0.065}{52}=0.00125$, then $1+\frac{0.065}{52}=1 + 0.00125=1.00125$

Step4: Calculate the exponent

$kt=52\times40 = 2080$

Step5: Calculate the value of $(1.00125)^{2080}$

Using a calculator, $(1.00125)^{2080}\approx14.77937$

Step6: Calculate the final amount A

$A = 100\times14.77937=1477.94$

Answer:

$1477.94$