Pairing from both ends: a closed form for 1 + 2 + ⋯ + n

What to assume they know

• Sequences and terms (Order is the point: a sequence carries position as well as value)
• Arithmetic progressions, and that the counting numbers are one (Common difference, and why the nth term is a + (n − 1)d)
• Triangular numbers as running totals of counting numbers (Order is the point: a sequence carries position as well as value)
• Rearranging an equation of the form 2S = k to find S
• Multiplying a two-digit number by the next one, and dividing by 2
• Area of a rectangle as rows times columns, for the picture argument

What they should be able to do

• Add the first ten counting numbers by pairing from both ends, without adding term by term
• Explain why every such pair has the same total
• Derive 2S = n(n + 1) by writing the sum forwards and backwards, and state the closed form for S
• Justify the same result from a rectangular array of two interlocking staircases
• Compute a stated triangular number from the closed form
• Find the total of a run of consecutive numbers that does not start at 1, by subtracting one closed form from another
• Count how many terms such a run contains
• Restate the result as an average multiplied by a count, and say when that version applies

Where it usually goes wrong

• "Halve first, then pair." Halving early is where the arithmetic goes wrong. Double the sum, pair the columns, and divide at the very end — that ordering is what makes the odd-n case behave, since 2S always has an even number of terms to pair.
• **"The pair total is n + 1 by coincidence."** It is forced. The first pair is 1 + n; moving inward adds 1 on the left and subtracts 1 on the right, so no pair can differ from the first. Show two or three columns changing in opposite directions.
• "This gives the total of any progression." It gives the total of the counting numbers only. The chapter derives nothing else — a general AP total is not stated anywhere in the chapter, summary included. What it does supply is enough to reach every printed question anyway, and the honest count of items needing more than §8.5 gives directly is one: Exercise Set 8.2 item 5, and even that yields to factoring out a 3. Item 7 of the same set is 1 + 2 + ⋯ + 25, which is S₂₅ itself, and End-of-Chapter problem 6 is a difference of two S values — the very move p. 185 demonstrates on the run from 25 to 58. So the route the chapter leaves you is to reduce a run to counting numbers, by factoring out a common multiplier or by subtracting one closed form from another, and that route is never blocked.
• "25 to 58 is 33 terms, since 58 − 25 = 33." It is 34. Both ends are included, so the count is 58 − 24. This is the same fence-post that put the (n − 1) in the nth-term formula, and it is worth naming as the same mistake.
• **"Subtract S₂₅ to remove everything below 25."** That removes 25 as well. You want S₂₄.
• "The average trick works for any list." It works when the terms are evenly spaced, so that the middle of the list coincides with the average of the ends. Āryabhaṭa's phrasing is a statement about progressions, not about arbitrary numbers.
• **"The formula only works when n is even, because you need to pair terms up."** The doubling is precisely what removes that worry. Try it on n = 7 and let the class see 2S = 7 × 8 come out clean.

Questions to check understanding

• Compute the total of the first n counting numbers for a stated n
• Compute the total of a run of consecutive numbers between two given bounds, and state how many terms it holds
• Find a stated triangular number from the closed form
• Find the smallest n for which the running total exceeds a given bound
• Total an AP whose terms are all multiples of one number, by factoring the multiplier out
• Express a given number as a total of consecutive counting numbers
• Explain, in words, why the pairing method works

Examples worth working on the board

Values marked verified are worked out here on the chapter's stated data; this book prints no answer key.

• The ten-term case, as printed (§8.5, p. 183). The total of 1 through 10 is written out, then written again in reverse, and the two lines are set one above the other. The page states that each column adds to 11, that doubling gives eleven added ten times, and hence 2S = 110 and S = 55.
• The general case, as printed (§8.5, p. 184). The forwards line and the backwards line are added to give 2S = n(n + 1), and the closed form n(n + 1)/2 follows. The notation Sₙ is introduced at the foot of the page.
• Fig. 8.5 (p. 184). Read off the printed page: a cream panel holding a rectangular array of circles six wide and seven deep, so 42 in all — that is the way round the drawing has it, and the printed 7 × 6 does not settle the orientation on its own. A stepped zigzag line divides them; the circles above it are green and those below are pink, and each colour forms a staircase of 1, 2, 3, 4, 5, 6 — reading down the seven rows, green runs 6, 5, 4, 3, 2, 1, 0 and pink runs 0, 1, 2, 3, 4, 5, 6. The line of type under the figure records 2 × (1 + 2 + 3 + 4 + 5 + 6) = 7 × 6. Verified: 1 + 2 + 3 + 4 + 5 + 6 = 21 and 7 × 6 = 42 = 2 × 21. The picture is the whole argument in one frame — two copies of the staircase fit together into a rectangle whose sides are n and n + 1.
• The picture extended (p. 184): the page then states 2 × (1 + 2 + ⋯ + 10) = 10 × 11 and, in general, 2 × (1 + 2 + ⋯ + n) = n × (n + 1). Note the shape of the rectangle for the ten-term case is 11 across and 10 down.
• Āryabhaṭa's statement, as cited (p. 185): the result is credited to the Āryabhaṭīya, Chapter 2, Verse 19, which the page says gives two ways of computing the total, the second being to take the average of the first and last terms and multiply by how many terms there are. Verified on the chapter's own ten-term case: the average of 1 and 10 is 5.5, and 5.5 × 10 = 55.
• Three totals to compute (the Think and Reflect on p. 185): S₂₀, S₅₀ and S₁₀₀₀. Verified: 210, 1275 and 500500. The last one is the payoff — a thousand-term addition done in two operations.
• A run that does not start at 1 (§8.5, p. 185). The total of 25 through 58 is computed as S₅₈ − S₂₄, printed as (58 × 59)/2 − (24 × 25)/2, which the page evaluates as 29 × 59 − 12 × 25 = 1711 − 300 = 1411. Verified. Note the subtracted index is 24, not 25 — the terms removed are 1 through 24, so that 25 survives. Verified: the run holds 58 − 24 = 34 terms, and its average is (25 + 58)/2 = 41.5, with 41.5 × 34 = 1411, which is a good independent check to run.
• Triangular numbers from the formula (the second Think and Reflect on p. 185, which states tₙ = n(n + 1)/2 for the triangular numbers and asks for the 10th, 17th and 80th). Verified: 55, 153, 3240.
• Two AP-sum exercises that need this section (Exercise Set 8.2, p. 186). Item 5: how many two-digit numbers are divisible by 3, and what do they total? Verified: they run 12, 15, …, 99, which is 30 numbers, and their total is 1665 — reachable from this section by factoring out the 3 and using 3 × (S₃₃ − S₃) = 3 × (561 − 6). Item 7: marbles laid in rows of 1, 2, 3 and so on up to 25 rows. Verified: 325, which is S₂₅ directly.
• Two end-of-chapter items (p. 195). Problem 9: the smallest n for which the total of the first n counting numbers passes 1,000. Verified: S₄₄ = 990 and S₄₅ = 1035, so n = 45. Problem 6, starred: all the ways of writing 100 as a total of consecutive counting numbers. Verified: 18 + 19 + 20 + 21 + 22, and 9 + 10 + ⋯ + 16, besides 100 standing alone. Both are solved by the one-total-minus-another move of section 9.

Figures to have open

• Fig. 8.5 redrawn as two staircases in contrasting colours that visibly slide together into a rectangle n wide and (n + 1) tall — six wide and seven tall in the worked instance, which is the printed figure's own orientation and not the transpose. This is the chapter's own figure (p. 184) and the argument of section 7 depends on the two staircases being the same shape rotated — redraw it as a clean schematic rather than reproducing the printed panel.
• The forwards-and-backwards pair of lines with bracketed columns, annotated. The chapter sets this in type only; as a figure it is the core of sections 2–5. Standard schematic.
• A number-line strip from 1 to 58 with 1–24 shaded out, for section 9. Standard schematic.
• No photograph or portrait is needed; the chapter prints none in this section.

Where this sits in the book

• Chapter 8, §8.5 Sum of the First n Natural Numbers, pp. 183–185. The ten-term worked case opens on p. 183; the general derivation and Fig. 8.5 are on p. 184; the Āryabhaṭa citation, the run from 25 to 58 and the triangular-number connection are on p. 185.
• Fig. 8.5, p. 184, with the equation line printed beneath it.
• Three Think and Reflect boxes belong here: whether the method extends to 1 + 2 + ⋯ + 100 (p. 184), the three totals S₂₀, S₅₀ and S₁₀₀₀ (p. 185), and the triangular-number formula with three terms to find (p. 185).
• Exercise Set 8.2, p. 186, items 5 and 7. Items 1–4 and 6 belong to Common difference, and why the nth term is a + (n − 1)d.
• End-of-Chapter Exercises, p. 195, problems 6 and 9. Both carry a printed asterisk — in that list every problem from 3 onwards does, and only 1 and 2 do not, so an asterisk marks nothing distinctive about either of these two.
• The summary, p. 196, restates n(n + 1)/2 in the triangular-number bullet and notes it is also the total of the counting numbers.
• Deliberate cross-reference outside the chapter: the Āryabhaṭīya is also cited in Chapter 3 of this book, for the approximation to π.