Exercise 7.1 answers: Binomial Theorem

Open NCERT p. 132Matches NCERT’s answer

1. Use (a + b)⁵ = ⁵C₀a⁵ + ⁵C₁a⁴b + ⁵C₂a³b² + ⁵C₃a²b³ + ⁵C₄ab⁴ + ⁵C₅b⁵, with a = 1 and b = −2x.
2. The binomial coefficients ⁵C₀ … ⁵C₅ are 1, 5, 10, 10, 5, 1.
3. Each power of a is 1, so every term is just its coefficient times the matching power of (−2x).
4. Powers of (−2x) alternate in sign: (−2x)¹=−2x, (−2x)²=4x², (−2x)³=−8x³, (−2x)⁴=16x⁴, (−2x)⁵=−32x⁵.
5. Multiply each coefficient by the matching power and add: 1 + 5(−2x) + 10(4x²) + 10(−8x³) + 5(16x⁴) + (−32x⁵).

Answer1 − 10x + 40x² − 80x³ + 80x⁴ − 32x⁵

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Open NCERT p. 132Matches NCERT’s answer

1. Use (a + b)⁵ with a = 2/x and b = −x/2.
2. The coefficients ⁵C₀ … ⁵C₅ are 1, 5, 10, 10, 5, 1.
3. Write the five powers of a: (2/x)⁵, (2/x)⁴, (2/x)³, (2/x)², (2/x)¹.
4. Write the matching powers of b: (−x/2)⁰ up to (−x/2)⁵, and multiply each pair with its coefficient.
5. Term by term: 32/x⁵ − 5·16/x⁴·x/2 + 10·8/x³·x²/4 − 10·4/x²·x³/8 + 5·2/x·x⁴/16 − x⁵/32.
6. Simplify each term: 32/x⁵ − 40/x³ + 20/x − 5x + 5x³/8 − x⁵/32.

Answer32/x⁵ − 40/x³ + 20/x − 5x + 5x³/8 − x⁵/32

Watch this explained “A worked instance”, 3:10 into Substituting particular values, and the coefficient identities that drop out

Open NCERT p. 132Matches NCERT’s answer

1. Use (a + b)⁶ with a = 2x and b = −3.
2. The coefficients ⁶C₀ … ⁶C₆ are 1, 6, 15, 20, 15, 6, 1.
3. Powers of a: (2x)⁶=64x⁶, (2x)⁵=32x⁵, (2x)⁴=16x⁴, (2x)³=8x³, (2x)²=4x², (2x)¹=2x.
4. Powers of b alternate in sign: 1, −3, 9, −27, 81, −243, 729.
5. Multiply each pair with its coefficient and add: 64x⁶ + 6(32x⁵)(−3) + 15(16x⁴)(9) + 20(8x³)(−27) + 15(4x²)(81) + 6(2x)(−243) + 729.

Answer64x⁶ − 576x⁵ + 2160x⁴ − 4320x³ + 4860x² − 2916x + 729

Watch this explained “A worked instance”, 3:10 into Substituting particular values, and the coefficient identities that drop out

Open NCERT p. 133Matches NCERT’s answer

1. Use (a + b)⁵ = ⁵C₀a⁵ + ⁵C₁a⁴b + ⁵C₂a³b² + ⁵C₃a²b³ + ⁵C₄ab⁴ + ⁵C₅b⁵ with a = x/3 and b = 1/x.
2. The coefficients ⁵C₀ … ⁵C₅ are 1, 5, 10, 10, 5, 1.
3. Write each term, raising the whole of x/3 and the whole of 1/x: (x/3)⁵ + 5(x/3)⁴(1/x) + 10(x/3)³(1/x)² + 10(x/3)²(1/x)³ + 5(x/3)(1/x)⁴ + (1/x)⁵.
4. Work out each term: (x/3)⁵ = x⁵/243; 5 · (x⁴/81) · (1/x) = 5x³/81; 10 · (x³/27) · (1/x²) = 10x/27; 10 · (x²/9) · (1/x³) = 10/(9x); 5 · (x/3) · (1/x⁴) = 5/(3x³); (1/x)⁵ = 1/x⁵.
5. Add the six terms: x⁵/243 + 5x³/81 + 10x/27 + 10/(9x) + 5/(3x³) + 1/x⁵.

Answerx⁵/243 + 5x³/81 + 10x/27 + 10/(9x) + 5/(3x³) + 1/x⁵

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

Open NCERT p. 133Matches NCERT’s answer

1. Use (a + b)⁶ = ⁶C₀a⁶ + ⁶C₁a⁵b + ⁶C₂a⁴b² + ⁶C₃a³b³ + ⁶C₄a²b⁴ + ⁶C₅ab⁵ + ⁶C₆b⁶ with a = x and b = 1/x.
2. The coefficients ⁶C₀ … ⁶C₆ are 1, 6, 15, 20, 15, 6, 1.
3. Write each term: x⁶ + 6x⁵(1/x) + 15x⁴(1/x²) + 20x³(1/x³) + 15x²(1/x⁴) + 6x(1/x⁵) + (1/x⁶).
4. Simplify each term by cancelling powers of x: x⁶, 6x⁴, 15x², 20, 15/x², 6/x⁴, 1/x⁶.
5. Add them: x⁶ + 6x⁴ + 15x² + 20 + 15/x² + 6/x⁴ + 1/x⁶.

Answerx⁶ + 6x⁴ + 15x² + 20 + 15/x² + 6/x⁴ + 1/x⁶

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. Write 96 as 100 − 4, so (96)³ = (100 − 4)³.
2. Use (a + b)³ = ³C₀a³ + ³C₁a²b + ³C₂ab² + ³C₃b³ with a = 100, b = −4.
3. The coefficients ³C₀ … ³C₃ are 1, 3, 3, 1.
4. Terms: (100)³ = 1000000; 3(100)²(−4) = −120000; 3(100)(16) = 4800; (−4)³ = −64.
5. Add the terms: 1000000 − 120000 + 4800 − 64 = 884736.

Answer884736

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. Write 102 as 100 + 2, so (102)⁵ = (100 + 2)⁵.
2. Use (a + b)⁵ = ⁵C₀a⁵ + ⁵C₁a⁴b + ⁵C₂a³b² + ⁵C₃a²b³ + ⁵C₄ab⁴ + ⁵C₅b⁵ with a = 100, b = 2.
3. The coefficients ⁵C₀ … ⁵C₅ are 1, 5, 10, 10, 5, 1.
4. Terms: 10000000000, 1000000000, 40000000, 800000, 8000, 32.
5. Add the terms: 10000000000 + 1000000000 + 40000000 + 800000 + 8000 + 32 = 11040808032.

Answer11040808032

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Open NCERT p. 133Matches NCERT’s answer

1. Write 101 as 100 + 1, so (101)⁴ = (100 + 1)⁴.
2. The coefficients ⁴C₀ … ⁴C₄ are 1, 4, 6, 4, 1.
3. Since every power of 1 is 1, the terms are just the coefficients times the powers of 100.
4. Terms: 100000000, 4000000, 60000, 400, 1.
5. Add the terms: 100000000 + 4000000 + 60000 + 400 + 1 = 104060401.

Answer104060401

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Open NCERT p. 133Matches NCERT’s answer

1. Write 99 as 100 − 1, so (99)⁵ = (100 − 1)⁵.
2. The coefficients ⁵C₀ … ⁵C₅ are 1, 5, 10, 10, 5, 1, and the powers of −1 alternate in sign.
3. Terms: 10000000000, −500000000, 10000000, −100000, 500, −1.
4. Add the terms: 10000000000 − 500000000 + 10000000 − 100000 + 500 − 1 = 9509900499.

Answer9509900499

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Open NCERT p. 133Checked by computer

1. Write (1.1)¹⁰⁰⁰⁰ as (1 + 0.1)¹⁰⁰⁰⁰.
2. By the Binomial Theorem, (1 + 0.1)¹⁰⁰⁰⁰ = ¹⁰⁰⁰⁰C₀ + ¹⁰⁰⁰⁰C₁(0.1) + other positive terms.
3. ¹⁰⁰⁰⁰C₀ = 1 and ¹⁰⁰⁰⁰C₁(0.1) = 10000 × 0.1 = 1000, so the sum of just these two terms is 1001.
4. Every other term in the expansion is positive, so the full value of (1.1)¹⁰⁰⁰⁰ is more than 1001.
5. Since 1001 is already bigger than 1000, (1.1)¹⁰⁰⁰⁰ is the larger number.

Answer(1.1)¹⁰⁰⁰⁰ is larger than 1000

Watch this explained “Bounding instead of computing”, 8:40 into Substituting particular values, and the coefficient identities that drop out

Open NCERT p. 133Matches NCERT’s answer

1. Expand (a + b)⁴ = a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴.
2. Expand (a − b)⁴ = a⁴ − 4a³b + 6a²b² − 4ab³ + b⁴.
3. Subtracting, every term with an even power of b cancels: (a + b)⁴ − (a − b)⁴ = 8a³b + 8ab³.
4. Now put a = √3, b = √2: 8a³b = 8·3√3·√2 = 24√6, and 8ab³ = 8·√3·2√2 = 16√6.
5. Add these: 24√6 + 16√6 = 40√6.

Answer(a + b)⁴ − (a − b)⁴ = 8a³b + 8ab³; so (√3 + √2)⁴ − (√3 − √2)⁴ = 40√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. Expand (x + 1)⁶ = x⁶ + 6x⁵ + 15x⁴ + 20x³ + 15x² + 6x + 1.
2. Expand (x − 1)⁶ = x⁶ − 6x⁵ + 15x⁴ − 20x³ + 15x² − 6x + 1.
3. Adding, every term with an odd power of x cancels: (x + 1)⁶ + (x − 1)⁶ = 2x⁶ + 30x⁴ + 30x² + 2.
4. Put x = √2: 2x⁶ = 2·8 = 16, 30x⁴ = 30·4 = 120, 30x² = 30·2 = 60, and the constant is 2.
5. Add these: 16 + 120 + 60 + 2 = 198.

Answer(x + 1)⁶ + (x − 1)⁶ = 2x⁶ + 30x⁴ + 30x² + 2; so (√2 + 1)⁶ + (√2 − 1)⁶ = 198

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Open NCERT p. 133One way to think about it

1. Prove: 9ⁿ⁺¹ − 8n − 9 is divisible by 64.
2. Write 9 = 1 + 8, so 9ⁿ⁺¹ = (1 + 8)ⁿ⁺¹.
3. By the Binomial Theorem, (1 + 8)ⁿ⁺¹ = ⁿ⁺¹C₀ + ⁿ⁺¹C₁(8) + ⁿ⁺¹C₂(8)² + … + ⁿ⁺¹Cₙ₊₁(8)ⁿ⁺¹.
4. The first two terms are ⁿ⁺¹C₀ = 1 and ⁿ⁺¹C₁(8) = 8(n + 1) = 8n + 8.
5. Every term from the third one onward has a factor of 8² = 64 in it, so 9ⁿ⁺¹ = 1 + 8n + 8 + 64k for some whole number k.
6. So 9ⁿ⁺¹ − 8n − 9 = (1 + 8n + 8 + 64k) − 8n − 9 = 64k, which is a multiple of 64.

In short9ⁿ⁺¹ − 8n − 9 always equals 64 times a whole number, so it is divisible by 64.

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

Open NCERT p. 133One way to think about it

1. Recall the Binomial Theorem: (a + b)ⁿ = ⁿC₀aⁿ + ⁿC₁aⁿ⁻¹b + … + ⁿCₙbⁿ = Σᵣ₌₀ⁿ ⁿCᵣ aⁿ⁻ʳ bʳ.
2. Put a = 1 and b = 3: (1 + 3)ⁿ = Σᵣ₌₀ⁿ ⁿCᵣ (1)ⁿ⁻ʳ (3)ʳ = Σᵣ₌₀ⁿ 3ʳ ⁿCᵣ, since every power of 1 is 1.
3. The left side is (1 + 3)ⁿ = 4ⁿ.
4. So Σᵣ₌₀ⁿ 3ʳ ⁿCᵣ = 4ⁿ, which is exactly what was to be shown.

In shortΣᵣ₌₀ⁿ 3ʳ ⁿCᵣ = 4ⁿ, shown by putting a = 1, b = 3 into the Binomial Theorem.

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