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