Suppose we wish to add the first $n$n terms of a geometric sequence. This is known as a series, or alternatively the partial sum of the sequence. We could write the sum as:
$S_n=a+ar+ar^2+ar^3+...+ar^{n-1}$Sn=a+ar+ar2+ar3+...+arn−1
If we multiply both sides of this equation by the common ratio $r$r we see that:
$rS_n=ar+ar^2+ar^3+...+ar^{n-1}+ar^n$rSn=ar+ar2+ar3+...+arn−1+arn
Then, by carefully subtracting $rS_n$rSn from $S_n$Sn term by term, we see that all of the middle terms disappear:
$S_n-rS_n=a+\left(ar-ar\right)+\left(ar^2-ar^2\right)+...+\left(ar^{n-1}-ar^{n-1}\right)-ar^n$Sn−rSn=a+(ar−ar)+(ar2−ar2)+...+(arn−1−arn−1)−arn
This means that $S_n-rS_n=a-ar^n$Sn−rSn=a−arn and when common factors are taken out on both sides of this equation, we find $S_n\left(1-r\right)=a\left(1-r^n\right)$Sn(1−r)=a(1−rn) . Finally, by dividing both sides by $\left(1-r\right)$(1−r) (excluding the trivial case of $r=1$r=1) we reveal the geometric sum formula:
$S_n=\frac{a\left(1-r^n\right)}{1-r}$Sn=a(1−rn)1−r
An extra step, multiplying the numerator and denominator by $-1$−1 , reveals a slightly different form for $S_n$Sn that is easier to manage when the common ratio is greater than $r=1$r=1 :
$S_n=\frac{a\left(r^n-1\right)}{r-1}$Sn=a(rn−1)r−1
As an example, adding up the first $10$10 terms of the geometric progression that begins $96+48+12+...$96+48+12+... is straightforward. Since $a=96$a=96 and $r=\frac{1}{2}$r=12 we have:
$S_{10}=\frac{96\times\left(1-\left(\frac{1}{2}\right)^{10}\right)}{1-\left(\frac{1}{2}\right)}=191.8125$S10=96×(1−(12)10)1−(12)=191.8125
To add up the first $20$20 terms of the sequence $2,6,18,...$2,6,18,... we might use the second version of the formula to reveal:
$S_{10}=\frac{2\times\left(3^{20}-1\right)}{3-1}=3486784400$S10=2×(320−1)3−1=3486784400
You might be wondering why the sum formulas exclude the case for $r=1$r=1. This is not an issue, for if $r=1$r=1, then the series becomes:
$S_n=a+a+a+...+a=a\times n$Sn=a+a+a+...+a=a×n
Consider the series $5+10+20$5+10+20 ...
Find the sum of the first $12$12 terms.
Consider the series $4-8+16-\text{. . . }-2048$4−8+16−. . . −2048.
Solve for $n$n, the number of terms in the series.
Find the sum of the series.
Consider the sum $0.25+0.0025+0.000025+\text{. . .}$0.25+0.0025+0.000025+. . .
Which decimal is the sum equivalent to?
$0.250250250$0.250250250$...$...
$0.252525$0.252525$...$...
$0.250250250$0.250250250$...$...
$0.252525$0.252525$...$...
Hence express $0.252525$0.252525$...$... as a fraction.
Suppose a rare species of frog can jump up to $2$2 meters in one bound. One such frog with an interest in mathematics sits $4$4 meters from a wall. The curious amphibian decides to jump $2$2 meters toward it in a single bound. After it completes the feat, it jumps again, but this time only a distance of $1$1 meter. Again the frog jumps, but only $\frac{1}{2}$12 a meter, then jumps again, and again, each time halving the distance it jumps. Will the frog get to the wall?
The total distance traveled toward the wall after the frog jumps $n$n times is given by $S_n=\frac{2\left(1-\left(\frac{1}{2}\right)^n\right)}{1-\frac{1}{2}}$Sn=2(1−(12)n)1−12 .
So after the tenth jump, the total distance is given by $S_{10}=4\left(1-\left(\frac{1}{2}\right)^{10}\right)=4\left(1-\frac{1}{1024}\right)$S10=4(1−(12)10)=4(1−11024)
If we look carefully at the last expression, we realize that, as the frog continues jumping toward the wall, the quantity inside the square parentheses approaches the value of $1$1 but will always be less than $1$1. This is the key observation that needs to be made. Therefore the entire sum must remain less than $4$4, but the frog can get as close to $4$4 as it likes simply by continuing to jump according to the geometric pattern described.
Whenever we have a geometric progression with its common ratio within the interval $-1
Since for any geometric series, $S_n=\frac{a\left(1-r^n\right)}{1-r}$Sn=a(1−rn)1−r , if the common ratio is within the interval $-1
$S_{\infty}=\frac{a}{1-r}$S∞=a1−r
Checking our frogs progress, $S_{\infty}=\frac{2}{1-\left(\frac{1}{2}\right)}=4$S∞=21−(12)=4 is the limiting sum.
As another example, the limiting sum of the geometric series $108+36+12\dots$108+36+12… is simply $S_{\infty}=\frac{108}{1-\left(\frac{1}{3}\right)}=162$S∞=1081−(13)=162.
Consider the infinite geometric sequence: $2$2, $\frac{1}{2}$12, $\frac{1}{8}$18, $\frac{1}{32}$132, $\ldots$…
Determine the common ratio, $r$r, between consecutive terms.
Find the limiting sum of the geometric series.
Consider the infinite geometric sequence: $125$125, $25$25, $5$5, $1$1, $\ldots$…
Determine the common ratio, $r$r, between consecutive terms.
Find the limiting sum of the geometric series.
Consider the infinite geometric sequence: $16$16, $-8$−8, $4$4, $-2$−2, $\ldots$…
Determine the common ratio between consecutive terms.
Find the limiting sum of the geometric series.
There are many real-life applications of geometric series. Formulas are often developed for many of these applications, particularly when they occur regularly in industry.
For our purposes, it is often best to find solutions to various problems starting from basic principles and the information given in the problem.
These examples will walk us through developing these formulas.
A bank client deposits $\$1000$$1000 at the beginning of each year, and is given $7%$7% interest per year for $50$50 years. How much will accrue in the account over that time?
Think: We might begin by searching for a pattern by examining what happens in the first few years.
Solve: If we set $A_n$An as the amount of money accrued after $n$n years have elapsed, then we have:
$A_0=1000$A0=1000
Then $A_1=1000+1000\times\frac{7}{100}=1000\left(1+\frac{7}{100}\right)=1000\times\left(1.07\right)^1$A1=1000+1000×7100=1000(1+7100)=1000×(1.07)1
This means that the amount accrued after $1$1 year becomes $\$1070$$1070.
By the end of the second year, another $\$1000$$1000 has been added, with interest, but the original $\$1000$$1000 has been boosted by two interest payments.
The total amount is determined as:
$A_2=1000\left(1.07\right)^1+\left[1000\left(1.07\right)\right]\times1.07=1000\left[1.07+1.07^2\right]$A2=1000(1.07)1+[1000(1.07)]×1.07=1000[1.07+1.072]
By the end of the third year, the total accrual becomes:
$A_3=1000\left[1.07+1.07^2+1.07^3\right]$A3=1000[1.07+1.072+1.073]
A pattern is emerging, and so by the end of $50$50 years, the total accrued becomes:
$A_{50}=1000\left[1.07+1.07^2+1.07^3+...+1.07^{50}\right]$A50=1000[1.07+1.072+1.073+...+1.0750].
Inside the square parentheses is a geometric series with first term and common ratio both equal to $1.07$1.07.
Recalling the formula for the sum of a geometric sequence as $S_n=\frac{a\left(r^n-1\right)}{r-1}$Sn=a(rn−1)r−1 we have for this series:
$A_n=1000\times\frac{1.07\left(1.07^{50}-1\right)}{1.07-1}=434985.96$An=1000×1.07(1.0750−1)1.07−1=434985.96
Hence the amount accrued in the account will be approximately $\$434986$$434986.
Reflect: Now let's devise a formula that a banker might use for any client wishing to deposit an amount $P$P at the beginning of ever year for $n$n years where an interest rate of $r%$r% per year is applied.
From our solution above, and calling $R=1+\frac{r}{100}$R=1+r100, we have the generalized formula given by:
$A_n$An | $=$= | $P\times\frac{R\left(R^n-1\right)}{R-1}$P×R(Rn−1)R−1 |
Show that the repeating decimal $N=0.2323232323...$N=0.2323232323... is a rational number. That is to say, the number can be put in the form $\frac{p}{q}$pq where $p$p and $q$q are integers and $q\ne0$q≠0.
Think: Can you we write the decimal as an infinite series?
The repeating decimal can be written: $N=0.23+0.0023+0.000023+...$N=0.23+0.0023+0.000023+...
The first term is given by $a=0.23$a=0.23 and the common ratio is $r=0.001$r=0.001.
Do: The sum becomes:
$N$N | $=$= | $\frac{a}{1-r}$a1−r |
$=$= | $\frac{0.23}{1-0.001}$0.231−0.001 | |
$=$= | $\frac{\frac{23}{100}}{1-\frac{1}{100}}$231001−1100 | |
$=$= | $\frac{\frac{23}{100}}{\frac{100-1}{100}}$23100100−1100 | |
$=$= | $\frac{23}{99}$2399 | |
Reflect: Can this same strategy can be applied to any repeating decimal?
The recipe for making a Koch snowflake is as follows;
The emerging figure has an infinite perimeter but we can show that it has a finite area. Develop a formula for that area.
Think: Can we use a series to describe the area?
Solve: The total Area $A_T$AT of the snow flake becomes:
$A_T$AT | $=$= | $A+3\left(\frac{A}{9}\right)+12\left(\frac{A}{9^2}\right)+48\left(\frac{A}{9^3}\right)+192\left(\frac{A}{9^4}\right)+...$A+3(A9)+12(A92)+48(A93)+192(A94)+... |
$=$= | $A\left[1+\left(\frac{3}{9}\right)+\left(\frac{12}{9^2}\right)+\left(\frac{48}{9^3}\right)+\left(\frac{192}{9^4}\right)+...\right]$A[1+(39)+(1292)+(4893)+(19294)+...] | |
The expression on the right and within the main bracket begins with the number $1$1, but all the other terms form a geometric series with first term $\frac{1}{3}$13 and common ratio $\frac{4}{9}$49. We can see this because the numerator of each fraction is increasing by a factor of $4$4 and the denominator is increasing by a factor of $9$9.
Since $|\frac{4}{9}|$|49| is less than $1$1, the series sums to $\frac{a}{1-r}=\frac{\frac{1}{3}}{1-\frac{4}{9}}=\frac{3}{5}$a1−r=131−49=35. Adding the extra $1$1 at the beginning, we see that the total sum within the main parentheses is $\frac{8}{5}$85.
This means that Koch snowflake has a total area given by $A_T=\frac{8A}{5}$AT=8A5.
Reflect: Now use your formula to find the area of a Koch snowflake whose equilateral triangles have an area $A=\frac{1}{3}$A=13.
The repeating decimal $0.8888\dots$0.8888… can be expressed as a fraction when viewed as an infinite geometric series.
Express the first decimal place, $0.8$0.8 as an unsimplified fraction.
Express the second decimal place, $0.08$0.08 as an unsimplified fraction.
Hence write, using fractions, the first five terms of the geometric sequence representing $0.8888\dots$0.8888…
State the values of $a$a, the first term, and $r$r, the common ratio, of this sequence.
$a$a$=$=$\editable{}$
$r$r$=$=$\editable{}$
If we add up infinitely many terms of this sequence, we will have the fraction equivalent of our repeating decimal. Calculate the infinite sum of the sequence as a fraction.
At the start of 2014 Pauline deposits $\$5000$$5000 into an investment account. At the end of each quarter she makes an extra deposit of $\$700$$700. By looking at the pattern investment, Pauline realizes she can use her knowledge of geometric series to find the balance in the account at some point in the future.
The table below shows the first few quarters of 2014. All values in the table are in dollars.
Quarter | Opening Balance | Interest | Deposit | Closing Balance |
---|---|---|---|---|
Jan-Mar | $5000$5000 | $200$200 | $700$700 | $5900$5900 |
Apr-Jun | $5900$5900 | $236.00$236.00 | $700$700 | $6836.00$6836.00 |
Jul-Sep | $6836.00$6836.00 | $273.44$273.44 | $700$700 | $7809.44$7809.44 |
Use the numbers for the January quarter to calculate the quarterly interest rate.
Write an expression for the amount in the account at the end of the first quarter.
Do not evaluate the expression.
$\editable{}\times\editable{}+\editable{}$×+
Using $5000\times1.04+700$5000×1.04+700 as the starting balance, write an expression for the amount in the account at the end of the second quarter.
$\editable{}\times\left(\editable{}\right)^2+\editable{}\times\editable{}+\editable{}$×()2+×+
Given that the amount in the account at the end of the second quarter can be expressed as $5000\times\left(1.04\right)^2+700\times1.04+700$5000×(1.04)2+700×1.04+700, write a similar expression for the amount in the account at the end of the third quarter.
$\editable{}\times\left(\editable{}\right)^3+\editable{}\times\left(\editable{}\right)^2+\editable{}\times\editable{}+\editable{}$×()3+×()2+×+
The amount in the account after $n$n quarters can be expressed as a term of a geometric sequence plus the sum of a geometric sequence.
Write an expression for the amount in the investment account after $n$n quarters.
Hence determine the total amount in Pauline’s account at the beginning of 2016 to the nearest dollar.
Rochelle invests $\$190000$$190000 at a rate of $7%$7% per annum compounded annually, and wants to work out how much she can withdraw each year to ensure the investment lasts $20$20 years.
We will use geometric sequences and series to determine what Rochelle's annual withdrawal amount should be if she wants the investment to last $20$20 years.
The amount in the account after $n$n years can be expressed as the $n$nth term of a geometric sequence minus the sum of a different geometric sequence.
Write an expression for the amount in the investment account after $n$n years.
Use $x$x to represent the amount to be withdrawn each year.
Hence determine Rochelle's annual withdrawal amount, correct to the nearest cent.
Derive the formula for the sum of a finite geometric series (when the common ratio is not 1), and use the formula to solve problems.