Cubes built from runs of consecutive odd numbers

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What to assume they know

Why the first n odd numbers add up to n² — that the first n odd numbers total n², and that the nth odd number is 2n − 1
What makes a number a perfect cube, and the three-identical-groups test — perfect cubes and the notation n³
Triangular numbers as running totals 1, 3, 6, 10, 15 (see Squares hiding inside triangular numbers)
The idea of an average of a set of numbers, at least informally
Adding a short run of consecutive odd numbers by hand, to check the shortcut

What they should be able to do

Read the chapter's printed pattern and state how many terms each run has
Explain why the runs use every odd number exactly once
Work out how many odd numbers have been used before a given run begins
Write down the first and last term of the nth run
Show that the run is symmetric about n²
Conclude that the run totals n × n², and check the conclusion against the printed runs
Total a long run without adding it, by identifying which run it is
Connect the two odd-number results — squares one term at a time, cubes a run at a time

Where it usually goes wrong

"The runs overlap, or skip some odd numbers." They do neither, and checking it is the first thing to do: run 3 ends at 11 and run 4 begins at 13.
**"Run n starts at the nth odd number."** Run 4 starts at 13, which is the seventh odd number. What has been spent already is a triangular number, not n.
"The run gives the gap between consecutive cubes." It gives the cube itself. Run 4 totals 64, not 64 − 27. Students who have just met the odd numbers as differences of squares reach for this reading immediately, so refuse it early.
"You have to add the ten numbers." The chapter asks for the total without the addition, and an explanation that adds them has answered a different question.
"The middle of the run is the average only when there is a middle." For an even-length run the average sits between the two central terms and is still n². Show one even case explicitly.
"Odd numbers make squares, so they cannot also make cubes." The same supply serves both. One taken at a time gives squares; taken in runs it gives cubes.
"The pattern is a coincidence in the first six lines." Six lines is data. The endpoint formula and the symmetry are the reason, and they hold for every n.

Questions to check understanding

Continue the printed pattern for the next run and state its total
Given a run of consecutive odd numbers, identify which cube it totals and say how you knew
Write down the first and last odd number in the run belonging to a stated cube
Total a long run without adding, and justify the shortcut
Decide whether a given odd number is the first term of some run
Explain in words why a run of n consecutive odd numbers centred on n² totals n³ — the reasoning form, where the computation earns no marks on its own

Examples worth working on the board

The six printed runs (Part I p.14), stepped down the page:
- 1 = 1 = 1³
- 3 + 5 = 8 = 2³
- 7 + 9 + 11 = 27 = 3³
- 13 + 15 + 17 + 19 = 64 = 4³
- 21 + 23 + 25 + 27 + 29 = 125 = 5³
- 31 + 33 + 35 + 37 + 39 + 41 = 216 = 6³
The run the chapter sets as the test (Part I p.14): 91 + 93 + 95 + 97 + 99 + 101 + 103 + 105 + 107 + 109. The question printed beneath asks for its total without carrying out the addition.
The structural facts the explanation needs — working added here on the chapter's printed runs, none of it stated on the page:
- Run n has n terms, and the runs take the odd numbers strictly in order, so the number of odd numbers used before run n starts is 1 + 2 + … + (n − 1), a triangular number.
- Since the kth odd number is 2k − 1, run n begins at n² − n + 1 and ends at n² + n − 1.
- Those two endpoints are the same distance either side of n², and the terms step by 2, so the run is symmetric about n². When n is odd, n² is literally the middle term; when n is even there is no middle term and n² sits between the two central ones.
- n terms averaging n² total n × n², which is n³.
Checks against the printed runs, arithmetic added here: run 5 should begin at 25 − 5 + 1 = 21 and end at 25 + 5 − 1 = 29, with 25 in the middle — the printed run is 21, 23, 25, 27, 29. Run 6 should begin at 36 − 6 + 1 = 31 and end at 36 + 6 − 1 = 41, with 36 falling between 35 and 37 — the printed run is 31, 33, 35, 37, 39, 41. Run 4 should begin at 13 and end at 19, with 16 between 15 and 17 — the printed run is 13, 15, 17, 19. All three match, which is the check section 7 should show.
Identifying the test run. Its first term is 91 and its last is 109. Solving n² − n + 1 = 91 and n² + n − 1 = 109 both give n = 10, so it is the tenth run: ten terms, symmetric about 10². Ten terms each averaging that centre gives the total in one multiplication — which is exactly the shortcut the chapter's question is fishing for.
The bonus for section 9 — an added derivation, and nowhere on the printed page. Runs 1 through n between them use the first 1 + 2 + … + n odd numbers. By the earlier result, the first k odd numbers total k². So 1³ + 2³ + … + n³ equals the square of 1 + 2 + … + n. Check on the printed runs: the first four runs total 1 + 8 + 27 + 64, and they consume the first ten odd numbers, whose total is 10² — and 1 + 2 + 3 + 4 is 10. This follows from two things the chapter does print and is a genuinely satisfying payoff; flag it as an extension rather than as something the book says.
Where it is used later (Part I p.17, item 5): 67³ − 66³ and 43³ − 42³ are set beside 67² − 66² and 43² − 42² and the student is asked which is greatest, with reasoning. That item belongs to Cube roots, and what successive differences expose, but the growth rate this topic makes visible is what settles it.

Figures to have open

A long tape of the odd numbers with scissor marks after positions 1, 3, 6, 10, 15, so the run lengths and the triangular cut points are visible at once. An added figure and the one that carries sections 2 and 3.
A single run drawn on a number line with its centre marked and the terms balanced either side, shown once for an odd-length run and once for an even-length one. An added figure.
The chapter's six printed runs, stepped as on the page (Part I p.14). Standard schematic.
No photograph is needed.

Where this sits in the book

NCERT Ganita Prakash Class 8, Part I, printed Chapter 1, §1.2 "Cubic Numbers", the subheading "Perfect Cubes and Consecutive Odd Numbers", Part I p.14, occupying the upper third of that page down to the "Cube Roots" subheading.
The square counterpart is §1.1's "Perfect Squares and Odd Numbers", Part I pp.5–7, covered by Why the first n odd numbers add up to n².
Triangular numbers, needed for section 3, are at Part I p.7; see Squares hiding inside triangular numbers.