Substituting particular values, and the coefficient identities that drop out

What to assume they know

• Proving the expansion for every positive power by induction — the theorem itself, its proof, and the range of powers it is claimed over
• Rewriting the triangle with selection counts, so any row is reachable directly — the coefficients as selection counts
• Laws of exponents applied to a negative quantity, so that even and odd powers can be told apart
• Division with quotient and remainder, and what it means for one number to divide another
• Simplifying surds far enough to see when two expressions differ only in a sign

What they should be able to do

• Derive the signed expansion by substituting the negative of the second term, and say which positions change sign and why
• Expand a bracket with a minus in it, keeping the signs and the compound terms straight
• Derive the form in which the first term is 1, and read the coefficients straight off it
• Show that a row of the triangle totals a power of two, by evaluating that form at a single value
• Show that the same row totals zero once alternate signs are attached, and state the powers for which that holds
• Evaluate a power of a two-digit number by splitting it about a round number
• Decide which of two quantities is larger using only the leading terms of an expansion, and justify discarding the rest
• Prove a remainder or divisibility claim that holds for every positive power, by substituting into the form whose first term is 1
• Explain why adding or subtracting two expansions that differ only in a sign wipes out half the terms, and use that to evaluate surd expressions

Where it usually goes wrong

• "The signed version is a second theorem." It is the first one with a negative quantity put in the second slot. Nothing was proved twice.
• "Signs alternate because subtraction alternates." They alternate because each position carries its own power of minus one, and even powers of a negative quantity are positive. A student who can say that will never lose a sign in the middle of a sixth power.
• "A row totalling a power of two is a separate fact about the triangle." It is the expansion read at a single point. The identity and the theorem are the same statement.
• "The alternating total is zero for every row." It is zero for the rows the theorem covers, which start at index 1. The single-entry row at the top totals 1, and the substitution shows why — it asks for a zeroth power of zero, which is the case p. 126 fenced off.
• "Example 3 needs the value of the power." It never computes it. Two terms give a total already above the bound and everything discarded is positive, so the comparison is settled. Students who try to evaluate the power have missed the method entirely.
• "Splitting a number for a binomial expansion is a trick with no rule." Split it about a round number whose powers you can write down, and keep the other part small so its high powers stay manageable. That is why 98 goes to 100 less 2 and not to 90 plus 8.
• "Example 4 proves a fact about 25 only." The same substitution proves the Exercise 7.1 item about 64, and the Miscellaneous item about a difference of like powers. The method is the content; the modulus is an input.
• "Adding two expansions that differ in a sign doubles everything." It doubles half of them and destroys the other half. Which half survives depends on whether you add or subtract, and the exercises deliberately ask for both.

Questions to check understanding

• Expand a bracket containing a minus sign, with compound or fractional terms
• Evaluate a power of a number near 100 or near 1 by splitting it and expanding
• Decide which of two quantities is larger using only the first two terms, with a justification for discarding the rest
• Approximate a power to a stated number of terms
• Prove a divisibility or remainder claim for every positive power by substituting into the unit-first-term expansion
• Evaluate a sum or difference of two surd expressions that differ only in a sign
• State and use the total of a row of the triangle, plain and with alternate signs
• Weighted-row identities in which a fixed number is substituted for the variable

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 signed expansion (§7.2.2 (i), p. 130). Input: the second term of the bracket replaced by the negative of a quantity. Verified: the r-th position picks up the r-th power of minus one, so positions 0, 2, 4 and so on keep their sign while positions 1, 3, 5 and so on reverse it. The page reaches the same conclusion by expanding first and simplifying afterwards, which is worth showing because students who memorise "signs alternate" cannot say where the alternation comes from.
• The printed instance (§7.2.2 (i), p. 130). Input: the bracket built from x and 2y, with a minus between them, raised to the fifth. Verified: the terms are x to the fifth, then 10x⁴y, then 40x³y², then 80x²y³, then 80xy⁴, then 32y⁵, with the second, fourth and sixth entering negative. The coefficients are the row 1, 5, 10, 10, 5, 1 multiplied by 1, 2, 4, 8, 16 and 32 in turn.
• The unit-first-term form (§7.2.2 (ii), p. 130). Input: the first term set to 1. Every power of it is 1, so the coefficients stand alone against successive powers of the remaining variable. This is the form every later application uses.
• The row total (p. 131). Input: the variable in that form set to 1. Verified: the left side becomes 2 raised to the power, and the right side is the bare sum of the row, so a row of the triangle totals a power of two. Check it against Fig 7.3 on p. 128: row 4 gives 1 + 4 + 6 + 4 + 1 = 16, and row 5 gives 1 + 5 + 10 + 10 + 5 + 1 = 32.
• The signed row total (§7.2.2 (iii), p. 131). Input: the first term set to 1 and the second to the negative of the variable, then the variable set to 1. Verified: the left side becomes zero raised to the power and the right side is the row with alternate signs attached, so that total is nothing. Check against Fig 7.3: row 4 gives 1 − 4 + 6 − 4 + 1 = 0, and row 5 gives 1 − 5 + 10 − 10 + 5 − 1 = 0.
• Where the two identities part company. Verified: the theorem is claimed for positive powers, and the doubling identity happens to survive at index 0 as well, since the single-entry row totals 1 and 2 to the power zero is 1. The signed identity does not survive there — that row totals 1, not 0 — and the reason is visible in the substitution, which asks for a zeroth power of zero. This is the same exclusion §7.2 flagged on p. 126 when it attached a condition to the zeroth power.
• Example 2 (p. 131). Input: the fifth power of 98, computed by writing 98 as 100 less 2. The page prints the two running totals it forms, 10040008000 from the positive positions and 1000800032 from the negative ones, and their difference, 9039207968. Verified: the six terms are 10000000000, then 1000000000, then 40000000, then 800000, then 8000, then 32, with the second, fourth and sixth subtracted, and the arithmetic closes exactly.
• Example 3 (pp. 131–132, beginning at the foot of p. 131). Input: a comparison between 1.01 raised to the millionth power and 10,000, settled by writing 1.01 as 1 plus one hundredth. Verified: the first two terms alone give 1 plus 10,000, and every remaining term is a positive number times a positive power, so nothing that follows can pull the total back down. The argument never computes the power; it bounds it. That is the transferable idea and Exercise 7.1 item 10 repeats it with 1.1 raised to the ten-thousandth power against 1000, where the first two terms give 1 plus 1000.
• Example 4 (p. 132). Input: the claim that six raised to any positive power, less five times that power, leaves 1 on division by 25. The page substitutes 5 into the unit-first-term form, so that six to the power appears as 1, plus five times the power, plus a tail every term of which carries at least two factors of 5. Verified: pulling 25 out of that tail leaves a whole-number factor built from the row's selection counts and rising powers of 5, so the difference is 25 times something, plus 1. Note the edge the page glosses over: at the first power the tail is empty and that factor is 0, so the claim holds there with nothing left to multiply.
• The cancelling pairs (Exercise 7.1 items 11 and 12, p. 133). Item 11: the difference between the fourth power of a sum and the fourth power of the corresponding difference, then the same with the two terms taken as the square roots of 3 and 2. Verified: the difference leaves twice the positions that changed sign, which is 8ab times the sum of the two squares, and the surd case gives 40 times the square root of 6. Item 12: the sum of the sixth powers of x plus 1 and x minus 1, then the same with x taken as the square root of 2. Verified: the sum leaves twice the even positions, which is 2(x⁶ + 15x⁴ + 15x² + 1), and the surd case gives 198.
• Numerical items (Exercise 7.1, p. 133). Item 6: the cube of 96. Item 7: the fifth power of 102. Item 8: the fourth power of 101. Item 9: the fifth power of 99. Verified by splitting each about 100: 884736; 11040808032; 104060401; 9509900499.
• Plain signed expansions (Exercise 7.1, p. 132). Item 1: 1 minus 2x, to the fifth. Item 2: 2 over x minus x over 2, to the fifth. Item 3: 2x minus 3, to the sixth. Verified for item 1: 1, then 10x, then 40x², then 80x³, then 80x⁴, then 32x⁵, alternating in sign — the same six coefficients as the worked instance in §7.2.2, with the roles of the two terms swapped. Verified for item 3: the seven magnitudes run 64x⁶, then 576x⁵, then 2160x⁴, then 4320x³, then 4860x², then 2916x, and a constant 729, with the signs alternating.
• Coefficient identities set as exercises. Exercise 7.1 item 13, p. 133: nine raised to one more than a positive power, less eight times that power, less nine, is divisible by 64. Verified: substituting 8 into the unit-first-term form makes the first two terms cancel against the two subtractions, and every survivor carries at least two factors of 8. Item 14, p. 133: the row of selection counts weighted by successive powers of 3 totals a power of 4. Verified: it is the unit-first-term form evaluated at 3, since 1 plus 3 is 4.
• Miscellaneous Exercise items (p. 133). Item 1: for distinct whole numbers, their difference divides the difference of their like powers; the printed hint rewrites the first number as its difference from the second, plus the second. Verified: every term of that expansion except the last carries the difference as a factor, and the last term is exactly what is being subtracted. Item 2: the difference of the sixth powers of the sum and the difference of the square roots of 3 and 2. Verified: 396 times the square root of 6. Item 3: the sum of the fourth powers of a square plus and minus the square root of one less than that square. Verified: 2a⁸ + 12a⁶ − 10a⁴ − 4a² + 2, the odd positions having cancelled. Item 4: an approximation to the fifth power of 0.99 from the first three terms only. Verified: 1, less 0.05, plus 0.001, giving 0.951.

Figures to have open

• A substitution tree: the theorem at the root, three branches for the three substitutions §7.2.2 makes, and the identities and examples hanging off each. Standard schematic, and the organising image of this topic.
• A sign-parity strip: positions along one expansion, each showing its power of minus one resolved to plus or minus. Standard schematic.
• A row-total panel built on Fig 7.3 from p. 128, showing one printed row summed twice — once plain, once with alternate signs. It reuses the chapter's own figure and should be drawn from it rather than reproduced.
• A banked-arithmetic panel for the fifth power of 98: six magnitudes in a column, positives on one side, negatives on the other, two totals and a difference. Standard schematic.
• A bounding panel for the comparison example: a number line with the bound marked, two computed terms already carrying the total past it, and the remaining terms drawn as arrows that can only point the same way. Standard schematic.
• No photograph is required.

Where this sits in the book

• NCERT Class XI Mathematics, Chapter 7 Binomial Theorem, §7.2.2, pp. 130–131, for the three substitutions, the worked fifth power with a minus sign, and the two row-total identities.
• Examples 2, 3 and 4, pp. 131–132, for the numerical power, the comparison and the remainder claim.
• Exercise 7.1, items 1, 2, 3, 6, 7, 8, 9, 10, 11, 12, 13 and 14, pp. 132–133.
• Miscellaneous Exercise on Chapter 7, items 1, 2, 3 and 4, p. 133.
• Backward pointer inside the same chapter: the theorem being substituted into is §7.2.1, p. 129, and the condition attached to the zeroth power is on p. 126.
• Fig 7.3, p. 128, supplies the printed rows used to check both row-total identities.