Luke deposits $3,500 into each of two savings accounts.
Account I earns 3% annual simple interest.
Account II 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 II at the end of 4 years?
For Account I, 3% of $3,500 is calculated as (3/100)*$3,500 = $<<3500*3/100=105>>105 in interest earned each year.
Thus, at the end of 4 years, the interest earned would be $105*4 = $<<105*4=420>>420.
The final balance for Account I would be $3,500 + $420 = $<<3500+420=3920>>3,920.
For Account II, the interest is compounded annually. After 4 years, the balance can be calculated using the formula A = P(1+r/n)^(nt), where A is the final balance, P is the principal amount, r is the annual interest rate (in decimal form), n is the number of times compounded per year, and t is the number of years.
In this case, P = $3,500, r = 3% (or 0.03), n = 1 (compounded annually), and t = 4 years. Plugging the values into the formula, we get:
A = $3,500(1+0.03/1)^(1*4)
A = $3,500(1+0.03)^4
A = $3,500(1.03)^4
A ≈ $3,683.02 (rounded to the nearest cent)
Therefore, at the end of 4 years, the sum of the balances for Account I and Account II is $3,920 + $3,683.02 = $<<3920+3683.02=7603.02>>7,603.02. Answer: \boxed{7603.02}.