Review of Geometric Series
a, ar, ar^{2}, ... , ar^{n-1}
is a GEOMETRIC SEQUENCE where a is the first term and r is the common ratio.
When these terms are summed or added together a GEOMETRIC SERIES is obtained.
Compound Interest
A useful application of the geometric sequence is for compound interest problems.
If an initial amount of money $P is invested at a compound interest rate of r % per period and if A_{1}, A_{2} .. A_{n} are the accumulated amounts at the end of each period then
A_{k + 1} = A_{k} + r/100. A_{k} = A_{k}(1 + r/100)
If A_{0 }= initial amount = P then we see that
A_{1} = P(1 + r/100)
A_{2} = A_{1}(1 + r/100) = P(1 + r/100)^{2}
Thus deduce that :
A_{n} = P(1 + r/100)^{n}
Examples
Example One
$1000 is invested for two years at an annual rate of interest of 8% payable monthly.
How much is the investment worth at the end of the two years?
Here the period is a month
Therefore, n = 2 × 12 = 24
r = 8/12 = 2/3Using the compound interest formula above
A_{24} = 1000(1 + 2/300)^{24}= $1173
Example Two
Suppose I deposit $100 at the start of each month into an account earning 7.5% per annum. Interest is compounded monthly. How much will I have in the account after 5 years?
Here we track the progress of each monthly deposit.
The first $100 is invested for 60 months and earns 7.5/12 % interest per month.
Therefore, Total = 100(1 + 7.5/1200)^{60} = A_{1}The second amount is invested for 59 weeks
Therefore, A_{2} = 100(1 + 7.5/1200)^{59} etc.Therefore, Total = A_{1} + A_{2} + ... A_{60}
= 100(1 + 7.5/1200)^{60} + ... + 100(1 + 7.5/1200)
= 100(1 + 7.5/1200)[(1 + 7.5/1200)^{59} + ... +1]
In the square brackets [ ] is a geometric series.
with a = 1 and r = (1 + 7.5/1200) and n = 60
= $7298
This example is typical of a superannuation fund type investment.
Time payments or Mortgages
A second application of geometric series to financial problems is in time payments ormortgages. They are also called "reducible interest loans" as interest is charged on a decreasing amount owed after each repayment.
Suppose $A is loaned to be repaid in equal instalments with an interest rate of r % per period being charged on money still owed.
Let the period repayment be $M
Let A_{1}, A_{2} ,,, etc. be the amount still owing after each period.
Then A_{1} = A(1 + r/100) − M
A_{2} = A_{1}(1 + r/100) − M
= A(1 + r/100)^{2} − M(1 + r/100) − M
Deduce A_{n} = A(1 + r/100)^{n} − M(1 + r/100)^{n -1} − ... M(1 + r/100) − M
But A_{n} = 0 since loan now repaid.
Therefore, A(1 + r/100)^{n} = M[(1 + r/100)^{n-1} + ... + (1 + r/100) + 1]
Again the [ ] contains a geometric series.
Note: Do not try to memorise this formula but work each problem following the steps of the derivation.
Example One
A $100 000 mortgage is to be repaid over 20 years in monthly instalments. The annual interest rate is 6.95% and is applied monthly.
Calculate the monthly repayments.
Here n = 20 × 12 = 240
r = 6.95/12
A = 100 000
Note: This means that total repayments are 240 × $772.3 = $185352.
Thus $85352 interest is paid!!!
Example Two
In the above example how much has been repaid after 10 years?
Let us calculate A_{120}
= 199970 − 133306
= $66664 still owing
i.e. $33336 repaid.
Proofs By Induction
Suppose we have some statement S(n) depending on the positive integer n.
Let S = set of all positive integers for which S(n) is true.
Suppose we can prove the following:
(i) 1 is a member of S i.e. S(1) is true.
(ii) If all the positive integers 1, 2, ..., k are members of S then so is k+1.
i.e. S(1), S(2), ...S(k) are true then so is S(k + 1)
Then we may deduce that S(n) is true for all positive integers.
This process is called proof by induction.
Example
Prove S(n):
Proof
Now assume S(1), S(2), ... S(k) are all true
specifically
Therefore, If we add (k+1) to both sides.
1 + 2 + ...+ k + (k + 1) = 1/2.k(k + 1) + (k + 1)
Therefore, S(k +1) is true
Therefore, The proof by induction is complete and
The proof works as follows:
If 1 S and (ii) has been proved then 1, 2 S. In turn this allows us to deduce 1, 2, 3 S and so on without limit. In most examples you only need the assumption S(k) is true to deduce S(k + 1).
Example One |
Example Two Prove 7^{n} + 2 is divisible by 3 for all positive integers. We can rewrite this as: Assume S(k): 7^{k} + 2 = 3m is true. Consider 7^{k + 1} + 2 = 7^{k}.7 + 2 Now if m is an integer so is 7m − 4 By induction 7^{n} + 2 is always divisible by 3 |
In the theory of Binomial Expansions the coefficient of x^{k} in the expansion (1 + x)^{n} is written ^{n}C_{k}
Therefore, (1 + x)^{n} = ^{n}C_{0} + ^{n}C_{1} x + ^{n}C_{2} x^{2} + ... + ^{n}C_{n} x^{n}
Putting x = 0 gives^{ n}C_{0} = 1
Also the Pascal's Triangle theorem states that:
^{n -1}C_{k -1} + ^{n -1}C_{k} = ^{n}C_{k}
This, together with induction can be used to prove
Therefore, true for n = 2
Assume the formula is true for all coefficients in the expansion of (1 + x)^{n − 1}
In particular:
And this is known to be true by Pascal's Triangle
Therefore, If true for all coefficients in (1 + x)n also true for all coefficients in (1 + x)^{n}But proved for n = 1 and n = 2
Therefore, true for all positive integers.
Note:
Induction can be extended for use in proving statements based on any set with
(i) An initial value
(ii) Every member has a "next" element
e.g. S = {4, 5, 6, ...} all integers ≥ 4
S = {1/4, 1/2, 3/4, ..} all multiples of 1/4 ≥ 1/4
Induction cannot be used on sets of real numbers since a real number has no "next" element.