compute 3000[(1.04)^12 + (1.04)^11 + ... + 1]=3000[((1.04)^13 -1)/ ((1.04) -1)]=$49,880.51.

other question:

have infinite series 6000[v + v^2 + v^3 + ...]=6000[v/(1-v)], where v is the discount rate 1/1.06.

get $100,000.00.

Jennifer is your oldest child. Tomorrow is her 5th birthday. You plan to set up a savings plan for Jennifer's university education. You will deposit $3,000, tomorrow and every year up to and including her 17th birthday. How much will be available to Jennifer on her 18th birthday? Assume the relevant interest rate is 4% per year compounded annually.