If the first term of a geometric progression is $t_1$t1 and the common ratio is $r$r, then the sequence becomes:
$t_1$t1, $t_1r$t1r, $t_1r^2$t1r2, $t_1r^3$t1r3, $t_1r^4$t1r4...
Suppose we wish to add the first $n$n terms of this sequence. This will form a geometric series, also known as the partial sum of the sequence. We could write the sum as:
$S_n=t_1+t_1r+t_1r^2+t_1r^3+...+t_1r^{n-1}$Sn=t1+t1r+t1r2+t1r3+...+t1rn−1
If we multiply both sides of this equation by the common ratio $r$r we see that:
$rS_n=t_1r+t_1r^2+t_1r^3+...+t_1r^{n-1}+t_1r^n$rSn=t1r+t1r2+t1r3+...+t1rn−1+t1rn
Then, by carefully subtracting $rS_n$rSn from $S_n$Sn term by term, we see that all of the middle terms disappear:
$S_n-rS_n=t_1+\left(t_1r-t_1r\right)+\left(t_1r^2-t_1r^2\right)+...+\left(t_1r^{n-1}-t_1r^{n-1}\right)-t_1r^n$Sn−rSn=t1+(t1r−t1r)+(t1r2−t1r2)+...+(t1rn−1−t1rn−1)−t1rn
This means that $S_n-rS_n=t_1-t_1r^n$Sn−rSn=t1−t1rn and when common factors are taken out on both sides of this equation, we find $S_n\left(1-r\right)=t_1\left(1-r^n\right)$Sn(1−r)=t1(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:
$S_n=\frac{t_1\left(1-r^n\right)}{1-r}$Sn=t1(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 :
$S_n=\frac{t_1\left(r^n-1\right)}{r-1}$Sn=t1(rn−1)r−1
For any geometric sequence with starting value $t_1$t1 and common ratio $r$r, we can find the sum of the first $n$n terms, using:
$S_n=\frac{t_1\left(1-r^n\right)}{1-r}$Sn=t1(1−rn)1−r, for $r<1$r<1 or $S_n=\frac{t_1\left(r^n-1\right)}{r-1}$Sn=t1(rn−1)r−1, convenient if $r>1$r>1
Alternatively, if the last term $t_n$tn is known:
$S_n=\frac{t_1-rt_n}{1-r}$Sn=t1−rtn1−r, for $r<1$r<1 or $S_n=\frac{rt_n-t_1}{r-1}$Sn=rtn−t1r−1, convenient if $r>1$r>1
You might be wondering that the formulas above exclude the case for $r=1$r=1. For the case where $r=1$r=1, then the sequence becomes:
$t_1,t_1,t_1,t_1,....$t1,t1,t1,t1,.... and hence, the sum of the first $n$n terms is:
| $S_n$Sn | $=$= | $t_1+t_1+t_1+...+t_1$t1+t1+t1+...+t1 | ($n$n times) |
| $=$= | $nt_1$nt1 |
Worked examples
Example 1
Find the sum of the first $10$10 terms of the geometric progression that begins $96+48+24+...$96+48+24+....
Think: State $t_1$t1, $r$r and $n$n, then substitute into the formula $S_n=\frac{t_1\left(1-r^n\right)}{1-r}$Sn=t1(1−rn)1−r, since we have a decreasing sequence $r$r must be less than $1$1.
Do: $t_1=96$t1=96, $r=\frac{1}{2}$r=12 and $n=10$n=10, so we have:
| $S_{10}$S10 | $=$= | $\frac{t_1\left(1-r^n\right)}{1-r}$t1(1−rn)1−r |
| $=$= | $\frac{96\times\left(1-\left(\frac{1}{2}\right)^{10}\right)}{1-\frac{1}{2}}$96×(1−(12)10)1−12 | |
| $=$= | $191.8125$191.8125 |
Example 2
Evaluate the geometric series $2+6+18+54...+4374$2+6+18+54...+4374.
Think: Here we know the fist and last term and have an increasing sequence. State $t_1$t1, $t_n$tn and $r$r, then substitute into the formula $S_n=\frac{rt_n-t_1}{r-1}$Sn=rtn−t1r−1. Note: we can use the formula $S_n=\frac{t_1\left(r^n-1\right)}{r-1}$Sn=t1(rn−1)r−1 but we would need to first solve $t_n=4374$tn=4374, for $n$n.
Do: $t_1=2$t1=2, $t_n=4374$tn=4374 and $r=3$r=3, so we have:
| $S_n$Sn | $=$= | $\frac{rt_n-t_1}{r-1}$rtn−t1r−1 |
| $=$= | $\frac{3\times4374-2}{3-1}$3×4374−23−1 | |
| $=$= | $6560$6560 |
Example 3
If the sum for the first $n$n terms of the geometric sequence $5,10,20,...$5,10,20,... is $5115$5115, find $n$n.
Think: We have an increasing geometric sequence. State $t_1$t1, $r$r and $S_n$Sn, then substitute into the formula $S_n=\frac{t_1\left(r^n-1\right)}{r-1}$Sn=t1(rn−1)r−1 and rearrange.
Do: $t_1=5$t1=5, $r=2$r=2 and $S_n=5115$Sn=5115, so we have:
| $S_n$Sn | $=$= | $\frac{t_1\left(r^n-1\right)}{r-1}$t1(rn−1)r−1 | |
| $5115$5115 | $=$= | $\frac{5\left(2^n-1\right)}{2-1}$5(2n−1)2−1 | Substitute values into formula |
| Hence,$5\left(2^n-1\right)$5(2n−1) | $=$= | $5115$5115 | Simplify fraction and bring unknown to left-hand side |
| $2^n-1$2n−1 | $=$= | $1023$1023 | Divide both sides by $5$5 |
| $2^n$2n | $=$= | $1024$1024 | Add $1$1 to both sides |
| $\therefore n$∴n | $=$= | $10$10 | Solve for $n$n, using guess and check, technology or logarithms. |
Example 4
A bank client deposits $\$1000$$1000 at the beginning of each year, and is given $7%$7% interest per year for $50$50 years. How much will accrue in the account over that time?
Think: In geometric sequences, we looked at finding the balance after $n$n years for a compound interest account, what makes this example different? Notice the client is also adding a deposit to the account each year. What is $t_1$t1 and $r$r in this case? To answer this, we might begin by searching for a pattern by examining what happens in the first few years - this is often a good starting point for application questions.
If we set $A_n$An as the amount of money accrued after $n$n years have elapsed, then we have:
$A_0=1000$A0=1000
And after one year we have:
| $A_1$A1 | $=$= | $1000+1000\times\frac{7}{100}$1000+1000×7100 |
| $=$= | $1000\left(1+\frac{7}{100}\right)$1000(1+7100) | |
| $=$= | $1000\times\left(1.07\right)$1000×(1.07) |
This means that the amount accrued after $1$1 year is $\$1070$$1070.
By the end of the second year, another $\$1000$$1000 has been added and receives interest, but the original $\$1000$$1000 has been boosted by two interest payments.
The total amount can be determined by:
| $A_2$A2 | $=$= | $1000\left(1.07\right)+1000\left(1.07\right)\left(1.07\right)$1000(1.07)+1000(1.07)(1.07) |
| $=$= | $1000\left(1.07+1.07^2\right)$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 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 brackets is a geometric series with first term and common ratio both equal to $1.07$1.07.
Hence, we can now state $t_1=1.07$t1=1.07, $r=1.07$r=1.07, $n=50$n=50 and substitute into the formula $A_n=1000\times S_n$An=1000×Sn, where $S_n=\frac{t_1\left(r^n-1\right)}{r-1}$Sn=t1(rn−1)r−1. So we have:
| $A_n$An | $=$= | $1000\left(\frac{t_1\left(r^n-1\right)}{r-1}\right)$1000(t1(rn−1)r−1) |
| $=$= | $1000\left(\frac{1.07\left(1.07^{50}-1\right)}{1.07-1}\right)$1000(1.07(1.0750−1)1.07−1) | |
| $\approx$≈ | $\$434985.95$$434985.95 |
Hence the amount accrued in the account will be approximately $\$434985.95$$434985.95.
Practice questions
Question 1
Consider the series $1+2+4$1+2+4 ...
Find the sum of the first $10$10 terms.
QUESTION 2
Consider the series $5-20+80-\text{. . . }-81920$5−20+80−. . . −81920.
Solve for $n$n, the number of terms in the series.
Find the sum of the series.
QUESTION 3
At the start of 2013 Laura 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, Laura realises 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 2013. All values in the table are in dollars.
| Quarter | Opening Balance | Interest | Deposit | Closing Balance |
|---|---|---|---|---|
| Jan-Mar | $5000$5000 | $150$150 | $700$700 | $5850$5850 |
| Apr-Jun | $5850$5850 | $175.50$175.50 | $700$700 | $6725.50$6725.50 |
| Jul-Sep | $6725.50$6725.50 | $201.77$201.77 | $700$700 | $7627.27$7627.27 |
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\times1.03+700$5000×1.03+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(1.03\right)^2+700\times1.03+700$5000×(1.03)2+700×1.03+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 Laura’s account at the beginning of 2015 to the nearest dollar.
Summation notation
Recall from arithmetic series we can write the series $2+4+6+8+10.....+30$2+4+6+8+10.....+30, using the following compact notation:
$\sum_{n=1}^{15}2n$15∑n=12n
The expression above can be read as "the sum of $2n$2n for the values $n=1$n=1 to $15$15".
Worked example
Example 5
Write the sum of the first $10$10 terms of the geometric sequence $4,12,36,...$4,12,36,... with summation notation and evaluate the expression.
Think: We want to sum from term $1$1 to term $10$10 for the sequence with $t_1=4$t1=4 and $r=3$r=3. This would have the general term of the form $t_n=4\left(3^{n-1}\right)$tn=4(3n−1). We could write this as: $\sum_{n=1}^{10}t_n$10∑n=1tn where $t_n=4\left(3^{n-1}\right)$tn=4(3n−1). Or we can write it is a single expression as:
$\sum_{n=1}^{10}4\left(3^{n-1}\right)$10∑n=14(3n−1)
Reflect: Substitute for $n=1,2,3...$n=1,2,3... the first few terms to see how the formula unfolds:
$4+12+36+...+78732$4+12+36+...+78732
To evaluate the summation we can use our geometric series formula: $S_n=\frac{t_1\left(r^n-1\right)}{r-1}$Sn=t1(rn−1)r−1:
| $S_{10}$S10 | $=$= | $\frac{4\left(3^{10}-1\right)}{3-1}$4(310−1)3−1 |
| $=$= | $118096$118096 |
Practice questions
Question 4
Consider the series $1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\text{. . .}+\frac{1}{1024}$1−12+14−18+. . .+11024.
Which of the following expresses the series in the form $\sum_{k=1}^{\editable{}}\editable{}$∑k=1?
$\sum_{k=1}^{11}\left(-\frac{1}{2}\right)^{k-1}$11∑k=1(−12)k−1
A$\sum_{k=0}^{10}\left(-\frac{1}{2}\right)^{k-1}$10∑k=0(−12)k−1
B$\sum_{k=1}^{11}\left(\frac{1}{2}\right)^{1-k}$11∑k=1(12)1−k
C$\sum_{k=1}^{11}\frac{1}{2^{k-1}}$11∑k=112k−1
D$-\sum_{k=1}^{11}\left(\frac{1}{2}\right)^{k-1}$−11∑k=1(12)k−1
E
Question 5
Using the properties of a geometric series, find $\sum_{j=1}^5243\left(\frac{5}{3}\right)^j$5∑j=1243(53)j.