HSC Mathematics Advanced, Extension 1, and Extension 2/Series and series applications
Sequences and Series
In primary school, you will likely remember dealing with sequences such as:
2, 4, 6, 8, ...
A common question, is how do we find the sum of sequence that consist of n terms without adding each term individually. In this section, we will go over the how to find the sum of a sequence.
Arithmetic Sequence
An arithmetic sequence is a sequence where each successive term increases or decreases by a fix amount.
Examples of Arithmetic Sequence
- In the sequence 3, 5, 7, 9, ... each successive term increases by 2.
- In the sequence 4, 1, -2, -5 each successive term decreases by 3. Notice that a sequence can have a finite number of terms. In this case, there are 4 terms. In the sequence 3, 5, 7, 9, ... there are an infinite number of terms as indicated by the "...".
Note that an arithmetic sequence with n terms can be written in the form a, a+d, a+2d, ..., a+(n-1)d. For our example involving 3, 5, 7, 9 notice that a=3, d=2. As a quick exercise you can verify through substitution the first 4 terms are 3, 5, 7, 9.
How do we know the sequence, a, a+d, a+2d, ..., a+(n-1)d has n terms?
Let start with some basic questions where we determine the number of terms in a sequence and work our way up to our original question.
Q: How many terms are there in the following sequences?
i) 1, 2, 3, 4, 5
ii) 0, 1, 2, 3, 4, 5
iii) 67, 68, 69, ..., 81, 82
iv) 0, 2, 4, 6, ..., 42
v) 7, 10, 13, ..., 28
Answers
i) There are 5 numbers. You can reach the answer through simply counting or performing the calculation 5-1+1
ii) There are 6 numbers. A common mistake is for people to perform the calculation 5-0=5. This fails to include the end point. For this reason, we perform the calculation 5-0+1 = 6. The +1 ensures we include the starting point.
iii) 82 - 67 + 1 = 16
iv) We divide each term by 2. We get the following sequence: 0, 1, 2, ..., 21. We can now find the number of terms similar to how we did ii) i.e. 21 - 0 + 1 = 22
v) Subtract each term in the sequence by 7. We get the following sequence: 0, 3, 6, ..., 21. Divide each term by 3. We get the following: 0, 1, 2, ..., 7. Performing the calculation 7-0+1 we find there are 8 terms in the sequence.
Applying the same method used in v) we can see there are n terms in the sequence a, a + d, a + 2d, ..., a + (n-1)d:
a, a+d, a+2d, ..., a+(n-1)d
Subtracting each term by, a we get
⇒ 0, d, 2d, ..., (n-1)d
Dividing each term by d we get
⇒ 0, 1, 2, ..., n-1
(n-1) - 0 + 1 = n. Therefore, the sequence has n terms.
Definition Arithmetic Sequence:
The n-term of an arithmetic sequence is a_n = a + (n-1)d. Alternatively the arithmetic sequence can be defined as a, a+d, a+2d, ..., a+(n-1)d.
Notice by defining the arithmetic sequence this way we avoid a case of "n" losing its meaning, i.e. a case where n does not equal the number of terms in the sequence. E.g., if we defined the arithmetic sequence as a, a+d, a+2d, ..., a+nd. There would n+1 terms in the sequence and not n.
How do we find the sum of an arithmetic sequence?
We can find the sum of an arithmetic sequence using the following theorem:
Theorem 1:
Suppose we have the sequence a, a+d, ..., a+(n-1)d i.e., an arithmetic sequence. There are two main ways to Then the sum of the sequence is
i) n(2a + (n-1)d)/2
which can be rewritten as
ii) (a+l)/2 * n where a is the first term of the sequence and l is the nth term of the sequence, i.e. l = a + (n-1)d.
Note: Theorem 1) ii) can be interpreted as the average of the first and last term in a sequence times by the number of terms is equal to the sum
Proof:
Let S represent the sum. Then:
S = (a) + (a+d) + (a+2d) + ... + (a+(n-1)d) (equation 1)
S = (a+(n-1)d) + a+(n-2)d + a+(n-3)d + ... + a (equation 2)
Notice how the terms are aligned in equation 1 and equation, i.e. a + a+(n-1)d = 2a + (n-1)d, (a+d) + a+(n-2)d = 2a + (n-1)d.
Therefore, adding equation 1 and equation 2 together, we get:
2S = 2a + (n-1)d + ... + 2a + (n-1)d
Where 2a + (n-1)d is added n times, since there are n terms in the arithmetic sequence.
Therefore, 2S = n(2a + (n-1)d)
⇒ S = n(2a + (n-1)d)/2
This proves Theorem 1 i).
We can rewrite n(2a + (n-1)d)/2 as n(a + a + (n-1)d)/2. We have defined l as being equal to the nth term of an arithmetic sequence which by definition is a + (n-1)d i.e. l = a + (n-1)d. Substituting l we get n(a + l)/2 = (a+l)/2*n as required.
Arithmetic Series Exercises
Find the sum of the following sequences:
i) 1, 2, 3, 4, ..., 100
ii) 41, 42, 43, ..., 69
iii) 0, 2, 4, ..., 12
iv) -1, -6, -11, ..., -46
v) 9, 15, 21, ..., 183
vi) 71, 68, 65, ..., -10
Geometric Sequence
In the previous section on arithmetic sequence, we look at sequences where each successive term increases or decreases by a fixed amount. In a geometric, each successive term is multiplied by some constant. An example of geometric sequence is:
i) 2, 6, 18, 54, ...
ii) 1, 1/2, 1/4, 1/8, ...
Note how in sequence i) each successive term is multiplied by 3. In sequence ii) each successive term is multiplied by 1/2.
Definition Geometric Sequence:
A geometric sequence has the form a, ar, ar^2, ..., ar^(n-1) where a is the first term of the sequence, r is the multiplicative constant and n represents the number of terms in the sequence.
We can also define the geometric sequence as a_n = ar^(n-1) where a_n is the nth term of the geometric sequence.
Why is nth term of the geometric series defined as a_n = ar^(n-1) and not a_n = ar^n?
Notice how there are n terms in the geometric sequence if a_n = ar^(n-1).
To see this, we can create a new sequence representing the power of the exponent. Since the power of r of the first term in seq 1) is 0 the first term in seq 2) is 0. Likewise, the power of r in the second term of seq 1) is 1 hence the second term of seq 2) is 1.
seq 1) a, ar, ar^2, ..., ar^(n-1)
seq 2) 0, 1, 2, ..., n-1
Using what we learned about the arithmetic sequence in the previous section, we can see the number of terms in seq 2) is:
(n-1) - 0+1 = n
Therefore, there are n terms in the geometric sequence. If we instead defined the terms as a_n = ar^n there would instead be n+1 terms in the sequence. This would mean n would lose its meaning and no longer correspond to the number terms in the sequence.
How can find the sum of a geometric sequence?
We can find the sum of a geometric sequence using the following theorem:
Theorem 2
Suppose we have the sequence a, ar, ..., ar^(n-1) i.e., a geometric sequence. Then the sum of the sequence is
a(1-r^n)/(1-r)
Proof
Let S = a + ar + ... + ar^(n-1). This means rS = ar + ar^2 + ... + ar^n. So we have two equations
S = a + ar + ... + ar^(n-1) (eq 1)
rS = ar + ar^2 + ... + ar^n (eq 2)
Subtracting eq 2 from eq 1 we get:
S - rS = [a + ar + ... + ar^(n-1)] - [ar + ar^2 + ... + ar^n]
=> S - rS = a - ar^n
=> S(1-r) = a(1-r^n)
=> S = a(1-r^n)/(1-r)
as required.
Geometric Series Exercises
Q1 Find the sum of the following sequence using theorem 2.
i) 3, 3^2, 3^3
ii) 3, 12, 48, ..., 49152
iii) 7, 49, ..., 13841287201
iv) 4, 4*6, ..., 4*6^5
Q2 Use the correct theorem based on the sequence
i) 11, 14, 17, ..., 65
ii) 2, 6, 18, ... (find the sum of the first 8 terms)
iii) 2, 8, 14, ... (find the sum of the first 10 terms)
iv) 7, 7*9, ..., 7*9^8
*Arithmetico-Geometric Sequence