As shown in Example, if a bank pays interest at a rate of

Chapter 5, Problem 22E

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As shown in Example, if a bank pays interest at a rate of i compounded m times a year, then the amount of money Pk at the end of k time periods (where one time period = 1 /mth of a year) satisfies the recurrence relation Pk = [1 + (i/m)] Pk-1 with initial condition P0 = the initial amount deposited. Find an explicit formula for Pn.ExampleCompound Interest with Compounding Several Times a YearWhen an annual interest rate of i is compounded m times per year, the interest rate paid per period is i/m. For instance, if 3% = 0.03 annual interest is compounded quarterly, then the interest rate paid per quarter is 0.03/4 = 0.0075.For each integer k ? 1, let Pk = the amount on deposit at the end of the kth period, assuming no additional deposits or withdrawals. Then the interest earned during the kth period equals the amount on deposit at the end of the (k –1)st period times the interest rate for the period: The amount on deposit at the end of the kth period, Pk , equals the amount at the end of the (k –1)st period, Pk–1, plus the interest earned during the kth period Suppose $10,000 is left on deposit at 3% compounded quarterly.a. How much will the account be worth at the end of one year, assuming no additional deposits or withdrawals?________________b. The annual percentage rate (APR) is the percentage increase in the value of the account over a one-year period. What is the APR for this account?Solutiona. For each integer n ? 1, let Pn= the amount on deposit after n consecutive quarters, assuming no additional deposits or withdrawals, and let P0 be the initial $10,000. Then by equation (5.5.4) with i = 0.03 and m = 4, a recurrence relation for the sequence P0, P1, P2,…is(1) Pk = Pk–1(1 + 0.0075) = (1.0075) • Pk–1 for all integers k? 1.The amount on deposit at the end of one year (four quarters), P4, can be found by successive substitution:(2) P0 = $10,000(3) P1 = 1.0075• P0 = (1.0075) $10,000.00 = $10,075.00 by (1) and (2)(4) P2 = 1.0075• P1 = (1.0075) $10,075.00 = $10,150.56 by (1) and (3)(5) P3 = 1.0075• P2 ? (1.0075)•$10,150.56 = $10,226.69 by (1) and (4)(6) P4 = 1.0075• P3? (1.0075) •$10,226.69 = $10,303.39 by (1) and (5)Hence after one year there is $10,303.39 (to the nearest cent) in the account.________________b. The percentage increase in the value of the account, or APR, is