Mathematics 451 Finding the Sum of Consecutive Terms of a Sequence In a previous handout we saw that an arithmetic sequence starts with an initial term b, and then each term is obtained by adding a common difference d to the previous term, so that the nth term is obtained by starting with the initial term and adding the common difference (n - 1) times. This is symbolized by: The Explicit Rule of an Arithmetic Sequence is given by f(n) = (initial term) + (n - 1)(common difference) or f(n) = b + (n - 1)d, where b is the initial term and d is the common difference. Suppose we want to add up the first few terms of such an arithmetic sequence. (Such a sum is called a finite arithmetic series.) A good problem solving strategy is to look at a simple case. The simplest sum of an arithmetic sequence is T(n) = 1 + 2 + 3 + 4 + 5 +... + n, which gives the nth triangular number; here b = 1 and d = 1. We can find a formula for T(n) by noticing that by pairing the first and last term we get a pair whose sum is n + 1; by pairing the second and next-to-last terms we get a pair whose sum is n + 1, and so on, working from the outside in. (This idea supposedly goes back to Gauss in the 1700s.) This strategy can be visualized by the diagram below, which shows the paired terms. 1 + 2 + 3 +.... (n-2)+(n-1)+n Since there are n terms, this leads to n/2 pairs, with each pair adding up to n + 1. Thus the entire sum is T(n) = 1 + 2 + 3 + 4 + 5 +... + n = (n/2)(n + 1) which is usually rewritten as T(n) = (n)(n+1)/2 This same strategy will work to find the sum of any arithmetic sequence. Suppose A = 1 + 3 + 5 + 7 + 9 +... + 95 + 97 + 99 (recursive: A n = 2 + A n-1 ) Sum of first and last terms of A: Sum of second and next-to-last terms of A: Sum of third and third-from-last terms of A: This leads us to: A can be grouped into pairs, each of which add up to. How many such pairs are there? Half the number of original terms! How many terms are there? You must first find the explicit description of the sequence. The common difference is d = 2, so the sequence is A(n) = 2(n - 1) + 1 = 2n - 1. What n gives 99? Set 99 = 2n - 1 and solve: 100 = 2n, so n = 50. Since A = 1 + 3 +... + 99 it is the sum of the first 50 terms of our sequence, so there are 50/2 = 25 pairs. Each pair sums to 100, so The sum 1 + 3 + 5 + 7 +... + 97 + 99 = 25 x 100 = 2500. In general this can be remembered as : Sum of an arithmetic sequence = (number of terms) x (sum of first and last terms) 2 Mathematics 451 Sums of Sequences Page 1
Let s see how this can be applied. 1. 2 + 5 + 8 + 11 +... + 101 =? The sum of the first and last terms are 2 + 101 = 103, that s easy. But how many terms are there? We need to know what term number 101 is in the sequence. To determine this we must work out the formula for the nth term of the sequence. We know a = 2 and d = 3 in this sequence, so the general formula in the box at the top of page 1 says that 101 = a + (n - 1) x d = 2 + (n - 1)x3 for this sequence. We solve this for n: 101 = 2 + 3n - 3 101 = 3n - 1 102 = 3n 34 = n There are 34 terms in our series, which we group into 17 pairs, each with sum 2 + 101 = 103. Thus the value (sum) of our series is 2 + 5 + 8 + 11 +... + 101 = 17 x 103 = 1751 Now you try a couple: 2. 12 + 17 + 22 + 27 +.... + 502 =? (recursive: next term = 5 + previous term) a) How many terms are there? b) Properly grouped, the sum of each pair of terms is c) So the sum of the series is 3. An auditorium has rows of seats, with 20 seats in the first row, 23 in the second row, 26 in the third, and so on in this arithmetic pattern to the last row which has 68 seats. a) How many rows are there in the auditorium? b) How many total seats are there in the auditorium? Recursive description: next row = a) Find the number of rows: b) Find the total number of seats: Mathematics 451 Sums of Sequences Page 2
Geometric Sequences Recall that to get the n th term of an geometric sequence, you start with the initial term and multiply the common ratio (n - 1) times. (The common ratio is often called the growth factor.) This is often stated as: The Explicit Rule for a Geometric Sequence is given by f(n) = (initial term) x (common ratio) (n - 1) or f(n) = b x r (n - 1), or an = b x r (n - 1) where b is the initial term and r is the common ratio (growth factor) An example of a geometric sequence is 2, 6, 18, 54, 162,..., where b = 2 and r = 3, whose rule is f(n) = 2x3 (n-1). When r is large compared to 1, as in this case, the size of the terms can grow quickly, in which case a term like the 10th might be very large. For the sequence just listed, f(10) = 2x3 9 = 39,366. (!) So finding the sum of the first 10 terms of this sequence would involve very large numbers! Consider the case is when r < 1, such as in the sequence 2, 2/3, 2/9, 2/27, 2/81,.. where a = 2 but now r = 1/3. The terms are getting smaller, but that will help keep the sum manageable. For this sequence, f(n) = 2x(1/3) (n-1) = 2/(3 (n-1) ) and f(10) = 2/(3 9 ) = 0.000101610... This means the sum of the first 9 terms will be almost the same as the sum of the first 10 terms, for their difference is f(10), which here is the small decimal shown above. In general, a finite geometric series is the sum of a collection of consecutive terms from a geometric sequence. Its general form is S = b + br + br 2 + br 3 +... + br (n-1) (**) where b is the initial term and r the common ratio How can we find a formula for this sum? Here is the standard trick : 1. Multiply both sides of (**) by r (raises the exponent on each r by 1), and subtract (**) from the result: r(**): rs = br + br 2 + br 3 + br 4 +...... + br (n) -(**): - S = -b - br - br 2 - br 3 -..... - br (n-1) Subtract: rs - S = -b + br (n) Notice that all but two of the terms on the right cancel, because all but two appear in both S and rs. 2. Now solve for S: Factor S out of the left hand side: S(r - 1) = br n -b Divide by (r - 1): S = Mathematics 451 Sums of Sequences Page 3
We conclude: If f(n) = br (n-1) gives the nth term of a geometric sequence, then the sum of the first n terms of the sequence is called a finite geometric series, and the value of this sum may be calculated as S = b + br + br 2 +.... + br (n-1) = = (term after last term) - (first term) r -1 Let s see how this can be applied. 1. 1 + 2 + 4 + 8 + 16 +... + 1024 =? (recursive description: next term = 2 x previous term) This is a geometric series. It is the sum of the first few terms in the geometric sequence 1, 2, 4, 8, 16,... This geometric sequence has common ratio (growth factor) r = 2 and first term b = 1. The next term in the sequence would be 2x1024 = 2048, so 2048 1 1 + 2 + 4 + 8 + 16 +... + 1024 = = 2047 2 1 Notes: (a) We do not need to determine which term 1024 is in the sequence we are summing. That is, we do not need to solve for n. (b) Consider the sequence of partial sums : 1, 1+2, 1+2+4, 1+2+4+8, 1+2+4+8+16, etc. This sequence of partial sums is 1, 3, 7, 15, 31,... (We ve seen this sequence before.) Its recursive description is: next term = (next power of 2) + previous term. The result 2047 above is the 10th term in this sequence of partial sums. Now you try a couple: 2. 1 + 3 2 + 9 4 + 27 8 +... + 6561 256 = S =? (Note: 38 = 6561, 2 8 = 256) Here r = and the next term is So S = 3. If you invest $1,000 per year at 6% annual compound interest for 40 years, what will be the total value of your investments? When a principal P is invested at an annual compound interest rate i for t years, the value of that single investment will be F(t) = P(1 + i) t Your first $1000 will earn interest for t = 40 years, so its value will be F(40) =1000(1+.06) 40 Your second $1000 will earn interest for t = 39 years, so its value will be F(39) =1000(1.06) 39 Your third $1000 will earn interest for t = 38 years, so its value will be F(38) =1000(1.06) 38 and so on. Your last $1000 will earn interest for t = 1 year, so its value will be F(1) =1000(1.06) 1 So the total value of your investments is S = 1000(1.06) + 1000(1.06) 2 +.... + 1000(1.06) 38 + 1000(1.06) 39 + 1000(1.06) 40 This is a geometric series with ratio r = between terms, so the total value may be calculated as S = (term after last term) - (first term) r -1 = Mathematics 451 Sums of Sequences Page 4
4. (a) Recall Leaping Leonora s contract #1, where she received a geometric sequence of salaries: $10,000, $20,000,..., $2,048,000 over ten years. Use the above method to quickly find her total payments over the 10 years: (b) Recall Leaping Leonora s contract #2, where she received a arithmetic sequence of salaries: $500,000, $1,000,000, $1,500,000... $5,000,000 Find her total payments using our technique for the sum of an arithmetic sequence: 5. Recall the meaning of a repeating decimal. For example, what does 0.7 mean? It means 0.77777... = 7 10 2 + 7 10 3 + 7 10 4... This last expression is the sum of an infinite geometric series. Its value is the limit of the sums of the finite geometric sequences 0.7 = 7 10, 0.77 = 7 10 2, 0.777 = 7 10 2 + 7, etc. In these finite sequences 3 10 the growth factor is r = 0.1 and the initial term is 0.7, so their values are 0.7(0.1) n 0.7 for n = 1, 2, 3,... What is the limit of as n? 0.1 1 Determine this limit, and you will find out the fraction that 0.7 represents. = 0.7(0.1)n 0.7 0.1 1 This method may be used to convert any repeating decimal to a fraction. Mathematics 451 Sums of Sequences Page 5