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University of Leeds MATH0380 FOUNDATION APPLIED MATHEMATICS FOR BUSINESS Handout: formula for monthly savings Reminder: A saver deposits an amount P in a savings account that pays r% interest compounded monthly and makes no further deposits. The amount in the account after n months is 1 r n Sn = P (1 + ) . 12 100 In this handout, we consider another way of saving. Instead of depositing one time and just waiting, the saver deposits the same amount every month into a savings account. The amount in the account after n months is n times the monthly deposit, plus any interest earned. We shall ﬁnd a formula for the total amount. The calculation is complicated because the money deposited earlier earns interest for a longer time than the money deposited later. Let us consider the example of a savings account that pays 6% interest compounded monthly. Suppose a saver deposits £100 on the ﬁrst day of every month. On the last day of every month, interest equal to 0.5% of the balance in the account is credited. Let In the amount in the account (in pounds) at the end of the nth month • After one month, the amount is I1 = 100(1.005). The deposit at the beginning of the month has earned one month’s interest. • After two months: I2 = (100(1.005) + 100) 1.005 = 100(1.005)2 + 100(1.005). (1) The ﬁrst deposit has earned interest twice; the second deposit has earned interest once. • After three months: I3 = 100(1.005)3 + 100(1.005)2 + 100(1.005). We can now write the formula for n months: In = 100(1.005)n + 100(1.005)n−1 + . . . + 100(1.005)2 + 100(1.005). (2) Instead of calculating the terms in (2) one by one using a calculator, it is more convenient to use a formula. To make it easier to write the formula, we use the notation 1 r p=1+ . (3) 12 100 Note that p is a number slightly greater than 1. In our example, r = 6 so p = 1.005. We can write the following relationship between the amount after n months and the amount after n + 1 months: In+1 = p(In + 100). (4) 1 Continued ... University of Leeds MATH0380 This says that, every month, £100 is added to the account and then the total amount of money in the account is multiplied by p. It is true for any n > 0. For example, choosing n = 1: I2 = p(I1 + 100) = 1.005(1.005 × 100 + 100) = 100(1.005)2 + 100(1.005), which is the same as (1). We can write a general formula for In . It is pn − 1 In = 100p . (5) p−1 How to ﬁnd the formula (5) First step: rewrite (2). In = 100 pn + pn−1 + . . . + p2 + p = 100 p pn−1 + pn−2 + . . . + p + 1 = 100 p Tn , (6) where we deﬁne Tn = pn−1 + pn−2 + . . . + p + 1. Second step: ﬁnd a formula for Tn by subtracting equations. pTn = pn + pn−1 + . . . + p2 + p Tn = pn−1 + pn−2 + . . . + p + 1 (subtract equations) p Tn − T n = pn − 1 (p − 1)Tn = pn − 1. Rearranging the last equation gives pn − 1 Tn = . (7) p−1 Third step: put (6) and (7) together. In = 100 p Tn pn − 1 = 100 p , p−1 which is what we wanted to prove. END 2

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posted: | 6/29/2009 |

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