Proving the expansion for every positive power by induction

What to assume they know

• Reading the pattern out of the first few expansions — the addition rule and the three regularities of §7.2
• Rewriting the triangle with selection counts, so any row is reachable directly — the coefficients written as selection counts, and Fig 7.3
• Choosing what to leave out, and the rule that builds each count from two smaller ones — the rule that builds one selection count from two smaller ones, which is the hinge of the whole proof
• Multiplying a many-term expression by a two-term bracket, and collecting like terms
• What it means to prove a statement for every member of an unbounded list, as against checking a few of them

What they should be able to do

• State the expansion §7.2.1 asserts, and name the two things it ranges over
• Say why checking the four indices from 1 to 4 does not establish the claim, and why the index-0 identity is not among the cases being checked
• Identify the base case the chapter uses, and explain why the argument starts at the first power rather than the zeroth
• Set out an induction hypothesis for this statement, assuming exactly one row
• Carry out the inductive step: multiply the assumed row by the bracket, identify the two copies produced, and align them
• Show that collecting like terms turns every interior coefficient into a sum of two adjacent selection counts
• Name the three small facts the page cites to finish the step, and say which of them does the real work
• State what an induction argument licenses and the exact claim it leaves outside
• Restate the theorem in the compact summation form, including the convention that makes the two end terms fit it
• Apply the theorem to a sixth power, and to a bracket one of whose terms is a fraction with a restriction attached

Where it usually goes wrong

• "It was checked for the first five powers, so it is true." No number of checked cases reaches an unbounded list. That gap is the reason §7.2.1 exists, and an explanation that skips the proof has taught the formula and none of the subject.
• "Induction is a ritual you perform after you already believe the result." The step is where the mathematics is. It shows that the addition rule of the triangle and the selection-count identity of Chapter 6 are the same statement, which is something a student cannot see from the table.
• "The proof begins at the zeroth power." It begins at the first. The zeroth power is not covered by the theorem as stated, and §7.2 gave it separately with a condition on the bracket.
• "The theorem covers any power at all." As stated here it covers positive whole powers. §7.1 says outright that other kinds of power are outside what this chapter treats, so an explanation must not promise fractional or negative powers.
• "The two copies produced in the step are different expansions." They are one expansion multiplied by each term of the bracket in turn. Their coefficients are the same list; only the alignment differs, and that offset is what makes adjacent entries meet.
• "The middle identity is proved here." It is imported. This chapter uses it and gives no argument for it; the argument belongs to Chapter 6.
• "The convention that a zeroth power reads as 1 is a technicality." Without it the compact summation form does not reproduce the two end terms, so it is what makes the short statement equivalent to the long one.
• "A three-term expression is outside the theorem." Group two of the three and it becomes a two-term bracket. Miscellaneous items 5 and 6 exist to make students choose the grouping.

Questions to check understanding

• State the theorem for a general positive power and identify the coefficient at a nominated position
• Reproduce the inductive step, naming at each line the fact that licenses it
• Explain why a finite check does not establish the theorem
• Expand a sixth or seventh power of a bracket with compound or fractional terms
• Rewrite a long expansion in the compact summation form, and back
• Group a three-term expression into a two-term bracket and expand it
• Short-answer items asking which powers the theorem as stated does and does not cover

Examples worth working on the board

Items marked verified are worked out here from the chapter's printed data; this chapter prints no answers on these pages.

• The statement (§7.2.1 heading, p. 129). A two-term bracket raised to a positive whole power equals a sum of terms, one per position along the row, in which the first term's index falls from the power to zero, the second term's index rises from zero to the power, and the coefficient at each position is the selection count for that row and position. The statement is set as the section heading, and the section heading is the citable form of it.
• The base case (§7.2.1, p. 129). Input: the first power. Verified: the two selection counts for row 1 are both 1, so the right-hand side collapses to the bracket itself and the two sides agree. The chapter starts here rather than at the zeroth power, and that choice is worth a beat of the explanation — the zeroth power is the one line of §7.2 that came with a condition attached.
• The hypothesis (§7.2.1, p. 129). The statement is assumed at one unnamed positive index, and nothing is assumed about any other index. The page labels this assumed equation and refers back to the label when it uses it, which is the discipline.
• The step (§7.2.1, p. 129), which is the content of this topic. Write the next power as the bracket times the assumed power. Substitute the assumed expansion. Distribute: the first term of the bracket multiplies every term of the assumed row, and so does the second. That produces two full copies of the row. In the first copy every first-term index has gone up by one; in the second every second-term index has gone up by one, which is the same list slid along by one position. Line the copies up and collect. Verified as a valid derivation: the page labels the multiplication step and the collecting step, and every fact it then cites was available before this chapter began.
• The three facts cited to finish (§7.2.1, p. 129). The selection count at position zero of the new row is 1; two adjacent selection counts of the old row add to the count sitting between and below them in the new row; and the count at the last position of the old row and of the new row are both 1. The middle one is the load-bearing fact and it is Chapter 6's, not this chapter's.
• The illustration (§7.2.1, p. 129). Input: the bracket built from x and 2, raised to the sixth power. Verified: the seven coefficients of row 6 are 1, 6, 15, 20, 15, 6, 1, and multiplying each by the matching power of 2 gives 1, 12, 60, 160, 240, 192 and 64. The page prints the same seven numbers. Note how fast the numerical coefficients grow away from the selection counts that produced them — the fifth of them, 240, is neither 15 nor 20.
• The five numbered observations (p. 130, printed under their own heading between §7.2.1 and §7.2.2). One: the compact summation form, with a counter running from zero to the power, together with the convention that a zeroth power of either term is read as 1 so that the two end terms fit the general shape. Two: the coefficients get their name here. Three: the count of terms exceeds the power by one. Four: the first term's index falls by one at each step from the power to zero, and the second term's index climbs by one from zero to the power. Five: in every term the two indices add to the power. Observations three, four and five restate as consequences of a proved theorem what §7.2 had offered as things noticed in a list.
• Example 1 (p. 131). Input: the bracket whose first term is x squared and whose second is 3 divided by x, raised to the fourth power, with x barred from being zero. Verified: the four powers of the second term bring in 3, 9, 27 and 81 while dividing by successive powers of x, and the five terms come out as x to the eighth, then 12x to the fifth, then 54x squared, then 108 divided by x, then 81 divided by x to the fourth. The restriction on x is content: the second term does not exist at zero, so the identity has nothing to say there.
• Exercise 7.1 items driven straight by the theorem (p. 133). Item 4: the bracket built from x over 3 and 1 over x, to the fifth. Item 5: the bracket built from x and 1 over x, to the sixth. Verified for item 5: the terms are x to the sixth, 6x to the fourth, 15x squared, 20, then 15, 6 and 1 divided by successive even powers of x — the constant 20 in the middle is the row-6 entry standing alone because the two indices cancel.
• Miscellaneous Exercise items 5 and 6 (p. 133). Item 5: an expression of three terms — 1, x over 2, and minus 2 over x — raised to the fourth, with x barred from zero. Item 6: an expression of three terms in x and a — 3x squared, minus 2ax, and 3a squared — raised to the third. Both are the theorem applied to something that is not a two-term bracket until you group two of the three terms together, and the grouping is a choice the theorem does not make for you.

Figures to have open

• A proof ladder: the statement at the top, the base case, the hypothesis, the distribution, the alignment, the merge and the conclusion as rungs, each with the licensing fact in a side column. Standard schematic, and the core image of this topic; the chapter prints the proof as running text with its labels in brackets.
• A two-copy alignment panel: one row of coefficients duplicated, the lower copy slid one position right, columns summing downward into the next row. This is the same schematic used in Reading the pattern out of the first few expansions, deliberately reused so the student sees the numerical and symbolic versions as one thing.
• An annotated pair of statements — the long expansion and the compact summation form — with lines joining the pieces that correspond. Standard schematic.
• A growth panel showing row 6 as selection counts beside the same expansion's numerical coefficients once the powers of 2 are folded in. Standard schematic.
• No textbook figure is required for this topic; §7.2.1 and the observations carry no numbered figure.

Where this sits in the book

• NCERT Class XI Mathematics, Chapter 7 Binomial Theorem, §7.2.1, p. 129, for the statement in the section heading, the proof, the three cited facts and the worked sixth power.
• The numbered observations, p. 130, for the compact summation form, the naming of the coefficients and the three consequences restated there.
• Example 1, p. 131, for a fourth power with a fractional second term and its restriction.
• Exercise 7.1, items 4 and 5, p. 133; Miscellaneous Exercise on Chapter 7, items 5 and 6, p. 133.
• Backward pointers outside this chapter: the selection count, its notation and the identity fusing two adjacent counts are Chapter 6, Permutations and Combinations, §6.4.