Group, reverse-and-add, shift-and-subtract, telescope — collapse any sum.
Adding a sequence one term at a time is honest but slow — and for an infinite tail, impossible. So we look at the shape of the general term and reach for the move that makes the whole sum fall in on itself. If the terms split into stacks we already know how to add, we group. If pairing the ends gives a constant total, we reverse and add — that is Gauss's trick made general. If each term is an arithmetic factor times a geometric one, we shift and subtract by the common ratio and a geometric series drops out. And if each term splits into a front piece minus the next term's front piece, the sum telescopes: neighbours cancel and only the two ends remain. Four moves, one habit — read the term, pick the tool.
Throughout, we lean on the two sum formulas we already proved and the geometric sum from 26.3. Keep these four building blocks on hand:
| Sum | Closed form | From |
|---|---|---|
| 1 + 2 + ⋯ + n | n(n+1)2 | arithmetic, d = 1 |
| 12 + 22 + ⋯ + n2 | n(n+1)(2n+1)6 | the square pyramid |
| a + (a+d) + ⋯ + aₙ | n(a₁+aₙ)/2 | arithmetic sum |
| a + aq + ⋯ + aqn−1 | a₁(1−qn)/(1−q) | geometric, q ≠ 1 |
The simplest move barely counts as a trick. The sum operator Σ just means "add them all," and addition lets you regroup freely: if the general term is a sum of pieces, you may add each piece separately and then add the totals. In symbols,
Σk=1n (bk + ck) = Σk=1n bk + Σk=1n ck.
Take a term that mixes an arithmetic part with a geometric part: ak = 2k + 3k. Neither sum alone is hard — we just split them apart and use the formula each one wants.
Split: Σ(2k + 3k) = 2·Σk + Σ3k.
The first stack is arithmetic: Σk=1n k = n(n+1)/2, so 2·Σk = n(n+1).
The second is geometric with a₁ = 3, q = 3: Σk=1n 3k = 3(3n − 1)/(3 − 1) = (3n+1 − 3)/2.
Add the totals: Σ(2k + 3k) = n(n+1) + (3n+1 − 3)/2. At n = 4 that is 4·5 + (243 − 3)/2 = 20 + 120 = 140 — and the direct term-by-term sum (2+3) + (4+9) + (6+27) + (8+81) = 5 + 13 + 33 + 89 = 140 checks out. ✓
Grouping only works because addition regroups. A product of two sequences does not split — Σ(bk·ck) ≠ (Σbk)(Σck). For a product like k·2k you need a different move (Section 3).
Pick a mixed term, then dial n. Watch the two stacks add, each summed by its own formula, and the total verified against a direct addition (green ✓).
The schoolboy Gauss is said to have summed 1 + 2 + ⋯ + 100 in seconds. His insight: write the sum forwards and backwards, then add them column by column. Each column pairs a term with its mirror, and every pair has the same total — so the whole double sum is just (that total) × (how many terms).
Every one of the n columns adds to n + 1, so 2S = n(n+1) and S = n(n+1)/2. The same picture works for any arithmetic sequence: pair a₁ with aₙ, a₂ with aₙ₋₁, and so on — each pair totals a₁ + aₙ, giving Sₙ = n(a₁ + aₙ)/2. (We drew this as a rectangle back in 26.2.)
In an arithmetic sequence, stepping one term forward adds d and one term backward subtracts d — so a forward term and its mirror always move in equal-and-opposite amounts. Their sum never changes: it stays locked at a₁ + aₙ for every column. A constant column height is exactly what makes the doubled sum a clean rectangle.
What about Σ k·2k = 1·2 + 2·4 + 3·8 + ⋯ + n·2n? Each term is an arithmetic factor k times a geometric factor 2k. Grouping fails (a product does not split), and the terms are not collinear, so reverse-and-add gives unequal columns. The trick that works copies the derivation of the geometric sum itself: multiply the whole sum by the common ratio q, slide the copy over by one slot, and subtract.
Write S and 2S = qS, lining the copy up under the term one position to its right:
Subtracting lined-up columns, each surviving piece is (k − (k−1))·2k = 2k — the awkward k-factor cancels and a plain geometric series falls out, with one leftover end term n·2n+1. So
−S = (21 + ⋯ + 2n) − n·2n+1 = (2n+1 − 2) − n·2n+1,
S = n·2n+1 − (2n+1 − 2) = (n − 1)·2n+1 + 2.
Check n = 3: 1·2 + 2·4 + 3·8 = 2 + 8 + 24 = 34, and (3−1)·24 + 2 = 2·16 + 2 = 34. ✓
Two traps. First, line the copy up by the power, not the index — qS shifts everything one slot right. Second, don't forget the leftover last term n·2n+1; it sits outside the cancelling band and must be carried to the end.
Dial n. The figure plots the partial sums Sₙ as green dots; the readout aligns S and 2S, shows the cancellation, and checks the closed form against a direct sum (green ✓).
The prettiest move. Suppose each term can be written as a front piece minus the next term's front piece: ak = f(k) − f(k+1). Then in the sum almost everything cancels — each −f(k+1) destroys the +f(k+1) in the next term — and only the very first front piece and the very last back piece survive. The sum collapses like a folding telescope.
The classic case is ak = 1/(k(k+1)). Partial fractions split it exactly:
1k(k+1) = 1k − 1k+1.
(Check by combining the right side over the common denominator: (k+1) − k over k(k+1) = 1/(k(k+1)). ✓) Now sum from k = 1 to n and watch the chain collapse:
Every interior fraction appears once with a + and once with a −, so they all annihilate. Only the leading 1 and the trailing −1/(n+1) are left:
Sₙ = 1 − 1/(n+1).
As n grows, 1/(n+1) → 0, so the partial sums creep up toward 1 without ever reaching it — at n = 10 the sum is 1 − 1/11 = 10/11 ≈ 0.909, at n = 100 it is 100/101 ≈ 0.990. A genuine limit, built by pure cancellation.
Dial n. Green dots are the partial sums Sₙ; they climb toward the green limit line at 1. The readout cancels the chain and checks 1 − 1/(n+1) against a direct sum (green ✓).
Telescoping only works if the split actually telescopes — the back piece of one term must equal the front piece of the next. And keep the surviving boundary terms: it is 1 − 1/(n+1), not just 1. The limit is 1, but for a finite sum you owe the −1/(n+1).
Every sum starts the same way: look at the general term and name its shape. That one read points straight at the tool.
| Shape of the term aₖ | Method | Example |
|---|---|---|
| arithmetic (constant difference) | sum formula | Σ(3k − 1) = n(a₁+aₙ)/2 |
| geometric (constant ratio) | sum formula | Σ 2k = 2n+1 − 2 |
| a sum of known stacks | grouping | Σ(2k + 3k) |
| arithmetic × geometric | shift-and-subtract | Σ k·2k |
| a fraction that splits | telescoping | Σ 1/(k(k+1)) |
| a sum of powers (k, k²) | grouping + formulas | Σ(k² + k) |
Σ(k² + k) is a sum of powers → group and use both pyramid formulas: n(n+1)(2n+1)/6 + n(n+1)/2 = n(n+1)(n+2)/3.
Σ 1/((2k−1)(2k+1)) is a splittable fraction → 1/(2k−1) − 1/(2k+1), all over 2, telescopes to ½(1 − 1/(2n+1)).
Σ (2k−1)·3k is arithmetic × geometric → shift-and-subtract by q = 3.
Notice how the strand fits together. An arithmetic or geometric term has its own closed form — those are the lessons just behind us. Everything in this lesson is a way to reduce an unfamiliar sum to those known ones: grouping splits into them, shift-and-subtract manufactures one, telescoping cancels down to two terms. And in 26.5 we will prove such a closed form holds for every n at once — by induction.
Four moves collapse a sum, and the term's shape tells you which to grab. Group sums of known stacks; reverse-and-add when pairing the ends gives a constant; shift-and-subtract for arithmetic × geometric; telescope a fraction that splits — and always carry the surviving boundary terms.
| Move | When | Engine | Result |
|---|---|---|---|
| Grouping | aₖ = bk + ck | Σ splits over + | Σbₖ + Σcₖ |
| Reverse-and-add | arithmetic | pair the ends | Sₙ = n(a₁+aₙ)/2 |
| Shift-and-subtract | arith × geom, e.g. k·2k | S − qS | (n−1)·2n+1 + 2 |
| Telescoping | 1/(k(k+1)) | 1/k − 1/(k+1) | 1 − 1/(n+1) → 1 |
The building blocks behind all four: 1 + 2 + ⋯ + n = n(n+1)/2, 1² + 2² + ⋯ + n² = n(n+1)(2n+1)/6, and the geometric sum a₁(1 − qn)/(1 − q). Reduce every new sum to these.
Sum 7 + 9 + 11 + ⋯ to n terms, i.e. Σk=1n (2k + 5). Find a closed form, then evaluate it at n = 10.
This is arithmetic with a₁ = 7, d = 2, so aₙ = 2n + 5. Reverse-and-add: Sₙ = n(a₁ + aₙ)/2 = n(7 + 2n + 5)/2 = n(2n + 12)/2 = n(n + 6). At n = 10: 10·16 = 160. (Direct: 7+9+⋯+25 = 160. ✓)
Use grouping to sum Σk=1n (k + 2k), then check it at n = 4.
Split into a known arithmetic stack and a known geometric stack: Σk + Σ2k = n(n+1)/2 + (2n+1 − 2). At n = 4: 4·5/2 + (32 − 2) = 10 + 30 = 40. (Direct: (1+2)+(2+4)+(3+8)+(4+16) = 3+6+11+20 = 40. ✓)
Show that 1/(k(k+1)) = 1/k − 1/(k+1), then telescope the sum to n = 4.
Combine the right side: 1/k − 1/(k+1) = (k+1 − k)/(k(k+1)) = 1/(k(k+1)). ✓ Summing to n = 4: (1 − ½) + (½ − ⅓) + (⅓ − ¼) + (¼ − ⅕) = 1 − ⅕ = 4/5. Everything between the ends cancels; in general Sₙ = 1 − 1/(n+1).
Use shift-and-subtract on Σk=1n k·3k (so q = 3). Find the closed form and check it at n = 3.
S = Σk·3k; 3S shifts each power up one. Subtract: S − 3S = (3 + 3² + ⋯ + 3n) − n·3n+1. The geometric part is (3n+1 − 3)/2, so −2S = (3n+1 − 3)/2 − n·3n+1, giving S = ((2n − 1)·3n+1 + 3)/4. At n = 3: (5·81 + 3)/4 = 408/4 = 102. (Direct: 3 + 18 + 81 = 102. ✓)
Telescope Σk=1n 1/((2k − 1)(2k + 1)). (Hint: 1/((2k−1)(2k+1)) = ½(1/(2k−1) − 1/(2k+1)).)
With the ½-factor split, the chain is ½[(1 − ⅓) + (⅓ − ⅕) + ⋯ + (1/(2n−1) − 1/(2n+1))]. All interiors cancel, leaving ½(1 − 1/(2n+1)) = ½ · 2n/(2n+1) = n/(2n + 1). At n = 3: 3/7 ≈ 0.4286. ✓ As n → ∞ the sum → ½.
Sum Σk=1n (k2 + k) by grouping with the two pyramid formulas, and simplify.
Σk² + Σk = n(n+1)(2n+1)/6 + n(n+1)/2. Factor n(n+1)/6: = n(n+1)/6 · [(2n+1) + 3] = n(n+1)(2n+4)/6 = n(n+1)(n+2)/3. At n = 5: 5·6·7/3 = 70. (Direct: 2+6+12+20+30 = 70. ✓)
Six questions to lock it in. Tap the answer you think is right.
This lesson develops fluency with finite series and the algebra of summation, touching HSA-SSE.B.4 (derive and use the formula for a finite geometric series, here reached by shift-and-subtract) and HSF-BF.A.1 / HSF-BF.A.2 (building functions and sequences and combining them — grouping is exactly the linearity of Σ). Telescoping previews the limit of a sequence (the partial sums approach 1), a first informal taste of convergence that calculus will make precise. The shared decision-table habit — read the term's shape, choose the method — supports MP7 (look for and make use of structure). A good prompt at home: before computing, ask "what shape is the general term, and which of the four moves does it invite?"