Fold an arithmetic series end to end and every pair adds to the same first-plus-last
Pair the ends
A series is the sum of the terms of a sequence. The terms of an arithmetic sequence go up by the same amount each time, and there is a quick way to add them.
Take 1 + 2 + 3 + … + 10. Pair the first term with the last: 1 + 10 = 11. Then pair the second with the second-last: 2 + 9 = 11. Then 3 + 8 = 11. Every pair makes the same 11, because moving one place inward adds 1 to the left number and takes 1 from the right number, so the total of the pair does not change.
Count the pairs
Ten terms make five pairs: 1 + 10, 2 + 9, 3 + 8, 4 + 7 and 5 + 6. Each pair is worth 11, so the sum is 5 × 11 = 55.
Here is the same idea drawn. Write the sum twice, once forward and once backward, and put the two copies side by side. Each of the 10 places holds 11, so the two copies add to 10 × 11 = 110, and one copy is half of that: 55.
Set n = 6 and slide the second staircase in
The gold staircase is 1 + 2 + … + 6. Slide in a second copy, turned upside down: the two fill a 6 by 7 rectangle of 42 blocks, so one staircase holds 42 ÷ 2 = 21.
The formula
Any arithmetic series pairs up the same way. As one end rises by the common difference, the other end falls by the same amount, so every pair is worth the first term plus the last. With n terms there are pairs, so the sum is .
If n is odd, the term in the middle has no partner. It lies halfway between the first and last terms, so it is worth exactly half a pair, and the formula still works: .
With first term a and last term l, the sum of n terms is . When the last term is not given, use l = a + (n − 1)d, where d is the common difference, as in The nth Term. Then the sum is .
Ten terms from 2 to 29
Find 2 + 5 + 8 + … + 29. First count the terms. They go up by 3, and from 2 to 29 is 27, which is 27 ÷ 3 = 9 steps of 3. Nine steps join ten terms, so there are 10 terms.
Ten terms make 10 ÷ 2 = 5 pairs, and each pair is worth 2 + 29 = 31. So the sum is 5 × 31 = 155.
The five pairs of 2 + 5 + 8 + … + 29. Each is worth 31, so the sum is 5 × 31 = 155.
Page numbers
In the next problem, a magazine has pages numbered 1 to 40. Their sum comes from the same pairing: 40 pages make 20 pairs, each worth 1 + 40 = 41.
Patterns and page numbers · One Sheet Torn from a 40-Page Magazine
QuestionA magazine has 40 pages, numbered from 1 to 40. One sheet of paper is torn out of it. The page numbers on the pages that are left add up to 749. What are the two page numbers on the torn-out sheet?
1.Pair the first page with the last, the second with the second to last, and so on: 1 + 40 = 41, 2 + 39 = 41, and there are 40 ÷ 2 = 20 pairs. All 40 pages add up to 20 × 41 = 820.
Pair the pages from both ends: 1 + 40, 2 + 39, ... twenty pairs of 41, so all 40 pages sum to 820. 2.The pages that are left add up to 749, so the torn sheet carried 820 − 749 = 71.
The remaining pages sum to 749, so the torn sheet carried 820 − 749 = 71. 3.A sheet has two consecutive page numbers. Draw the front page as one unit, u, and the back page as one unit and 1 more.
A sheet has two consecutive page numbers: the front is one unit, the back one unit and 1 more. 4.Together they make 2u + 1 = 71, so 2u = 70 and u = 35.
2u + 1 = 71, so 2u = 70 and u = 35. 5.The torn sheet is pages 35 and 36. Check: 820 − 35 − 36 = 749.
The torn sheet is pages 35 and 36.
Answer: Pages 35 and 36
Watch out for
- Halving 71 and getting 35.5. The two pages are whole numbers 1 apart, so take the 1 away first: 71 − 1 = 70, and the front page is 70 ÷ 2 = 35.
- Adding the 40 pages as 40 × 40 = 1600 or 40 × 41 = 1640. Pairing the pages gives 20 pairs of 41, which is half of 40 × 41: 820.
- Taking only one page away. A sheet of paper has two sides, so tearing out one sheet removes two page numbers.
A series in sigma notation
A sum written with is often an arithmetic series. The sum has the terms 24, 26, 28, …, 82, which go up by 2. It can be added with . It can also be split into two simpler sums: 22 added once for each of the 30 values of r, and 2 times . By the pairing, is .
Sequences and Series · The Seats in a Stadium Stand Written in Sigma Notation, and the Seats Behind the Reserved Rows
QuestionA stand at a stadium has 30 rows of seats. Row r, counted from the front, has (22 + 2r) seats, so the front row has 24 seats. (a) Write the total number of seats in the stand in sigma notation, and evaluate it. (b) Rows 1 to 10 are reserved for season-ticket holders. Write the number of seats in rows 11 to 30 in sigma notation, and evaluate it.
1.Row r has (22 + 2r) seats, and r runs from 1 to 30. So the total is ∑r=130 (22 + 2r).
The total is ∑r=130 (22 + 2r): one term for each row. 2.Split the sum into ∑r=130 22 + 2∑r=130 r. The first part adds 22 once for each of the 30 rows, which is 30 × 22 = 660. The second part is 2 × 30 × 312 = 930.
Split it: ∑r=130 22 = 660 and 2∑r=130 r = 2 × 465 = 930. 3.(a) The stand has 660 + 930 = 1590 seats. Check with the first and last rows: row 30 has 22 + 60 = 82 seats, and 302(24 + 82) = 15 × 106 = 1590.
(a) The stand has 1590 seats. 4.The seats behind the reserved rows are ∑r=1130 (22 + 2r). This is the whole sum less rows 1 to 10, and ∑r=110 (22 + 2r) = 10 × 22 + 2 × 10 × 112 = 220 + 110 = 330.
The gold rows are ∑r=1130 (22 + 2r): the whole sum less ∑r=110 (22 + 2r) = 330. 5.(b) ∑r=1130 (22 + 2r) = 1590 − 330 = 1260 seats. Check: rows 11 to 30 are 20 rows, from 44 seats up to 82 seats, and 202(44 + 82) = 10 × 126 = 1260.
(b) Rows 11 to 30 have 1590 − 330 = 1260 seats.
Answer: (a) ∑r=130 (22 + 2r) = 1590 seats; (b) ∑r=1130 (22 + 2r) = 1260 seats
Watch out for
- Evaluating ∑r=130 22 as 22. The 22 is added once for every value of r, so it contributes 30 × 22 = 660.
- Taking away the first 11 rows instead of the first 10. The sum from row 11 to row 30 includes row 11, so the rows taken away are 1 to 10. Counting the rows that are left, 30 − 10 = 20, catches the slip.
Your turn
1.Folding 1 + 2 + … + 10 works because…