Luke deposits $3,500 into each of two savings accounts.
• Account I earns 3% annual simple interest.
• Account |I earns 3% interest compounded annually.
Luke does not make any additional deposits or withdrawals. What is the sum of the balances of Account I and Account Il at the end of 4 years?
For Account I, the interest is simple interest, so it does not compound over the years. After 4 years, the balance in Account I will be $3,500 * 3% * 4 = $<<3500*3*.01*4=420>>420.
For Account II, the interest is compounded annually. The formula for compound interest is A = P(1 + r/n)^(nt), where:
A = the future value of the investment/loan, including interest
P = the principal investment amount (the initial deposit or loan amount)
r = the annual interest rate (as a decimal)
n = the number of times that interest is compounded per year
t = the number of years the money is invested/borrowed for
In this case, P = $3,500, r = 3%, n = 1 (compounded annually), and t = 4.
So, for Account II, the balance after 4 years will be: $3,500 * (1 + 3%/1)^(1*4) = $<<3500*(1+3*.01*1)^(1*4)=3723.5>>3,723.50.
The sum of the balances in Account I and Account II will be $420 + $3,723.50 = $<<420+3723.5=4143.5>>4,143.50. Answer: \boxed{4,143.50}.
And you as sure that this is the correct answer?
Apologies for the confusion. Let's re-evaluate the solution.
For Account I, since it earns 3% annual simple interest, after 4 years, the balance in Account I will be:
$3,500 + ($3,500 * 3% * 4) = $3,500 + ($3,500 * 0.03 * 4) = $3,500 + $420 = $3,920.
For Account II, it earns 3% interest compounded annually, so the balance after 4 years will be:
$3,500 * (1 + 3%)^4 = $3,500 * (1 + 0.03)^4 = $3,500 * 1.03^4 = $3,500 * 1.126749 = $3,955.62.
The sum of the balances in Account I and Account II will be $3,920 + $3,955.62 = $7,875.62.
Therefore, the correct answer is $7,875.62.