b1. F = Pert , which assumes continuous compounding, says that the Future value (F) of an amount (P) invested today at an annual rate (r), expressed as a decimal for the time (t), in years is given by the function.  Thus if you invested $100 at the annual rate of 5 1/2% for 6 years and 3 months you would get back (at the end of the time), F = $100e(0.055)(6.25) = $100e(0.3438) = $100(1.4102) = $141.02.   If you invest $15000 today, what amount does the formula say you will get back if you leave it for 5 years and 3 months in a savings account paying 4 1/2% annually?

The force of interest measured in years is δ(t) = 0.02 + 0.01t for t ≥ 0. (a)Compute the accumulated value of £100 invested for 3 years. (b)Compute the constant effective rate per annum that in 3 years produces the same accumulated value of £100 as in (a).

Task: Assume that at time 0 a sum L is lent for a series of n yearly payments. The rth payment, of amount x, is due at the end of the rth year. Let the effective annual interest rate for the rth year be i,. Give an identity which expresses L in terms of the x, and i,. Answer: The identity is [ Select ] [ Select ] L = x_1 (1+i_1)^(-1) + x_2 (1+i_1)^(-1) (1+i_2)^(-2) + .. + x_n (1+i_1)^(-1) (1+i_2)^(-2) ... (1+i_n)^(-n) L = x_1 (1+i_1) + x_2 (1+i_1) (1+i_2) + ... + x_n (1+i_1) (1+i_2) ... (1+i_n) L = x_1 (1+i_1)^(-1) + x_2 (1+i_1)^(-1) (1+i_2)^(-1) + .. + x_n (1+i_1)^(-1) (1+i_2)^(-1) ... (1+i_n)^(-1) Question 3 L = x_1 (1+i_1) + x_2 (1+i_1) (1+i_2)^2 + ... + x_n (1+i_1) (1+i_2)^2 ... (1+i_n)^n

(Solving for n with non-annual periods) Approximately how many years would it take for an investment to grow sixfold if it were invested at 9 percent compounded weekly? Assume that you invest $1 today. If you invest $1 at 9 percent compounded weekly, about how many years would it take for your investment to grow sixfold to $6? (Hint: Remember to convert your calculator solution to years.) years (Round to one decimal place.)