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Savings Interest Formula

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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



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