Miscellaneous Exercise answers: Binomial Theorem

Open NCERT p. 133One way to think about it

1. Prove: a − b is a factor of aⁿ − bⁿ.
2. Follow the hint: a = (a − b) + b, so aⁿ = [(a − b) + b]ⁿ.
3. Expand by the Binomial Theorem, with (a − b) as the first term and b as the second: aⁿ = ⁿC₀(a − b)ⁿ + ⁿC₁(a − b)ⁿ⁻¹b + ⁿC₂(a − b)ⁿ⁻²b² + … + ⁿCₙ₋₁(a − b)bⁿ⁻¹ + ⁿCₙbⁿ.
4. The last term is ⁿCₙbⁿ = bⁿ. Take it to the left side: aⁿ − bⁿ = ⁿC₀(a − b)ⁿ + ⁿC₁(a − b)ⁿ⁻¹b + … + ⁿCₙ₋₁(a − b)bⁿ⁻¹.
5. In each term left on the right, (a − b) is raised to a power of at least 1 (the powers run from n down to 1), so (a − b) comes out of every term: aⁿ − bⁿ = (a − b)[ⁿC₀(a − b)ⁿ⁻¹ + ⁿC₁(a − b)ⁿ⁻²b + … + ⁿCₙ₋₁bⁿ⁻¹].
6. Call the bracket k. Every ⁿCᵣ is a whole number and a, b are integers, so k is an integer. Because a and b are distinct, a − b is not 0.
7. So aⁿ − bⁿ = (a − b)k with k an integer, which means a − b is a factor of aⁿ − bⁿ.

In shortaⁿ − bⁿ = (a − b)k, where k = ⁿC₀(a − b)ⁿ⁻¹ + ⁿC₁(a − b)ⁿ⁻²b + … + ⁿCₙ₋₁bⁿ⁻¹ is an integer, so a − b is a factor of aⁿ − bⁿ.

Watch this explained “A remainder at every power”, 9:48 into Substituting particular values, and the coefficient identities that drop out

Open NCERT p. 133Matches NCERT’s answer

1. Expand (a + b)⁶ and (a − b)⁶ with a = √3, b = √2.
2. When you subtract the second expansion from the first, every term with an even power of b cancels, leaving only the odd-power terms doubled.
3. The surviving terms are 2[⁶C₁a⁵b + ⁶C₃a³b³ + ⁶C₅ab⁵].
4. Compute each: ⁶C₁a⁵b = 6·9√3·√2 = 54√6; ⁶C₃a³b³ = 20·3√3·2√2 = 120√6; ⁶C₅ab⁵ = 6·√3·4√2 = 24√6.
5. Add and double: 2(54√6 + 120√6 + 24√6) = 2(198√6) = 396√6.

Answer396√6

Watch this explained “Pairs that cancel”, 11:07 into Substituting particular values, and the coefficient identities that drop out

Open NCERT p. 133Matches NCERT’s answer

1. Let p = a² and q = √(a² − 1), so the expression is (p + q)⁴ + (p − q)⁴.
2. Adding the two expansions cancels every odd-power-of-q term, leaving 2[⁴C₀p⁴ + ⁴C₂p²q² + ⁴C₄q⁴].
3. So the sum is 2p⁴ + 12p²q² + 2q⁴.
4. Substitute p = a² and q² = a² − 1: 2a⁸ + 12a⁴(a² − 1) + 2(a² − 1)².
5. Expand and collect: 2a⁸ + 12a⁶ − 12a⁴ + 2a⁴ − 4a² + 2 = 2a⁸ + 12a⁶ − 10a⁴ − 4a² + 2.

Answer2a⁸ + 12a⁶ − 10a⁴ − 4a² + 2

Watch this explained “Pairs that cancel”, 11:07 into Substituting particular values, and the coefficient identities that drop out

Open NCERT p. 133Matches NCERT’s answer

1. Write 0.99 as 1 − 0.01, so (0.99)⁵ = (1 − 0.01)⁵.
2. By the Binomial Theorem, (1 − 0.01)⁵ = ⁵C₀ − ⁵C₁(0.01) + ⁵C₂(0.01)² − … .
3. Take only the first three terms: 1 − 5(0.01) + 10(0.01)².
4. Compute each: 1, −0.05, and 10 × 0.0001 = 0.001.
5. Add them: 1 − 0.05 + 0.001 = 0.951.

Answer0.951

Watch this explained “A hard multiplication made easy”, 7:32 into Substituting particular values, and the coefficient identities that drop out

Open NCERT p. 133Matches NCERT’s answer

1. Group the bracket as two terms: (1 + x/2 − 2/x)⁴ = [(1 + x/2) − 2/x]⁴, and use ⁴C₀ … ⁴C₄ = 1, 4, 6, 4, 1.
2. So it equals (1 + x/2)⁴ − 4(1 + x/2)³(2/x) + 6(1 + x/2)²(2/x)² − 4(1 + x/2)(2/x)³ + (2/x)⁴.
3. Expand the inner brackets by the Binomial Theorem: (1 + x/2)⁴ = 1 + 2x + 3x²/2 + x³/2 + x⁴/16; (1 + x/2)³ = 1 + 3x/2 + 3x²/4 + x³/8; (1 + x/2)² = 1 + x + x²/4.
4. First term: 1 + 2x + 3x²/2 + x³/2 + x⁴/16.
5. Second term: −(8/x)(1 + 3x/2 + 3x²/4 + x³/8) = −8/x − 12 − 6x − x².
6. Third term: 6 · (4/x²)(1 + x + x²/4) = 24/x² + 24/x + 6.
7. Fourth term: −4 · (8/x³)(1 + x/2) = −32/x³ − 16/x².
8. Fifth term: (2/x)⁴ = 16/x⁴.
9. Collect like powers: x⁴/16; x³/2; 3x²/2 − x² = x²/2; 2x − 6x = −4x; 1 − 12 + 6 = −5; −8/x + 24/x = 16/x; 24/x² − 16/x² = 8/x²; −32/x³; 16/x⁴.

Answerx⁴/16 + x³/2 + x²/2 − 4x − 5 + 16/x + 8/x² − 32/x³ + 16/x⁴

Watch this explained “In use”, 11:07 into Proving the expansion for every positive power by induction

Open NCERT p. 133Matches NCERT’s answer

1. Group the bracket as two terms: (3x² − 2ax + 3a²)³ = [(3x² − 2ax) + 3a²]³, and use ³C₀ … ³C₃ = 1, 3, 3, 1.
2. So it equals (3x² − 2ax)³ + 3(3x² − 2ax)²(3a²) + 3(3x² − 2ax)(3a²)² + (3a²)³.
3. Expand the inner brackets by the Binomial Theorem: (3x² − 2ax)³ = 27x⁶ − 3(9x⁴)(2ax) + 3(3x²)(4a²x²) − 8a³x³ = 27x⁶ − 54ax⁵ + 36a²x⁴ − 8a³x³, and (3x² − 2ax)² = 9x⁴ − 12ax³ + 4a²x².
4. Second term: 3 · 3a² · (9x⁴ − 12ax³ + 4a²x²) = 9a²(9x⁴ − 12ax³ + 4a²x²) = 81a²x⁴ − 108a³x³ + 36a⁴x².
5. Third term: 3 · 9a⁴ · (3x² − 2ax) = 27a⁴(3x² − 2ax) = 81a⁴x² − 54a⁵x.
6. Fourth term: (3a²)³ = 27a⁶.
7. Collect like terms: 27x⁶; −54ax⁵; 36a²x⁴ + 81a²x⁴ = 117a²x⁴; −8a³x³ − 108a³x³ = −116a³x³; 36a⁴x² + 81a⁴x² = 117a⁴x²; −54a⁵x; 27a⁶.

Answer27x⁶ − 54ax⁵ + 117a²x⁴ − 116a³x³ + 117a⁴x² − 54a⁵x + 27a⁶

Watch this explained “Putting the letters back”, 10:04 into Reading the pattern out of the first few expansions