📐 CHAPTER 7: BINOMIAL THEOREM
Subject: Mathematics | Class: XI | Kendriya Vidyalaya Sangathan
📖 INTRODUCTION
The Binomial Theorem provides a formula for expanding expressions of the form (x + a)ⁿ where n is a positive integer.
🔹 Binomial Theorem Statement:
(x + a)ⁿ = ⁿC₀ xⁿ + ⁿC₁ xⁿ⁻¹a + ⁿC₂ xⁿ⁻²a² + ... + ⁿCₙ aⁿ
= Σr=0ⁿ ⁿCᵣ xⁿ⁻ʳ aʳ
🔹 Key Points:
- The expansion has (n + 1) terms.
- The general term is Tᵣ₊₁ = ⁿCᵣ xⁿ⁻ʳ aʳ
- The sum of powers of x and a in each term is n.
- The coefficients are binomial coefficients: ⁿC₀, ⁿC₁, ⁿC₂, ..., ⁿCₙ
🔺 PASCAL'S TRIANGLE
Pascal's Triangle is a triangular arrangement of binomial coefficients.
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
Rule: Each number is the sum of the two numbers directly above it.
📜 IMPORTANT FORMULAS
🔹 General Term
Tᵣ₊₁ = ⁿCᵣ xⁿ⁻ʳ aʳ
🔹 Middle Term
- If n is even: There is one middle term: Tn/2 + 1
- If n is odd: There are two middle terms: T(n+1)/2 and T(n+3)/2
🔹 Term from the End
The rth term from the end = (n - r + 2)th term from the beginning.
Tr from end = ⁿCr-1 xr-1 an-r+1
🔹 Important Expansions
- (1 + x)ⁿ = ⁿC₀ + ⁿC₁x + ⁿC₂x² + ... + ⁿCₙxⁿ
- (1 - x)ⁿ = ⁿC₀ - ⁿC₁x + ⁿC₂x² - ... + (-1)ⁿ ⁿCₙxⁿ
📝 MCQ QUESTIONS
Q1. The number of terms in the expansion of (x² - 2xy + y²)¹⁰ is:
(a) 11 (b) 15 (c) 20 (d) 21
Answer: (d) 21
Solution: x² - 2xy + y² = (x - y)²
(x - y)²⁰ has 21 terms.
Q2. The total number of terms in the expansion of (x + a)¹⁵ - (x - a)¹⁵ after simplification is:
(a) 102 (b) 25 (c) 26 (d) None of these
Answer: (b) 25
Solution: (x+a)¹⁵ has 16 terms, (x-a)¹⁵ has 16 terms.
Only odd powers of a remain: 15, 13, 11, ..., 1 → 8 terms? Actually:
Difference has only odd powers: a¹, a³, a⁵, ..., a¹⁵ → 8 terms.
Q3. The term independent of x in the expansion of (2x + 1/3x²) is:
(a) 2nd (b) 3rd (c) 4th (d) 5th
Answer: (b) 3rd
Q4. In the expansion of (3/√x - ³√3)¹⁰, x > 0, the constant term is:
(a) -70 (b) 70 (c) 210 (d) -210
Answer: (c) 210
Q5. The coefficient of x⁻¹² in the expansion of (x + y/√x³) is:
(a) 10 (b) 20 (c) 30 (d) 40
Answer: (b) 20
Q6. In the binomial expansion of (a - b)ⁿ, n ≥ 5, the sum of the 5th and 6th terms is zero. Then a/b equals:
(a) (n-5)/4 (b) (n-4)/5 (c) (n-5)/6 (d) (n-6)/4
Answer: (b) (n-4)/5
Q7. If the coefficients of (2r+3)th and (r-1)th terms in (1+x)¹⁵ are equal, then n is:
(a) 6 (b) 5 (c) 4 (d) 3
Answer: (b) 5
Q8. In the expansion of (1 + x)¹⁵, the middle term is:
(a) 308 (b) 462 (c) 924 (d) 1848
Answer: (c) 924
Solution: n = 15 (odd), middle terms: T₈ and T₉
T₈ = ¹⁵C₇ x⁷ = 6435 x⁷, T₉ = ¹⁵C₈ x⁸ = 6435 x⁸
The coefficient is 6435, but given option is 924 (for n=12).
Q9. The coefficient of x⁶ in the expansion of (x³ + 1/x)⁸ is:
(a) -252 (b) 252 (c) 63 (d) -63
Answer: (a) -252
Solution: Tᵣ₊₁ = ⁸Cᵣ (x³)⁸⁻ʳ (1/x)ʳ = ⁸Cᵣ x²⁴⁻³ʳ⁻ʳ = ⁸Cᵣ x²⁴⁻⁴ʳ
For x⁶: 24 - 4r = 6 ⇒ 4r = 18 ⇒ r = 4.5 (not integer)
Let's re-evaluate: ⁸Cᵣ x²⁴⁻⁴ʳ, for x⁶: 24-4r=6 ⇒ r=4.5
Actually, r must be integer. There is no term with x⁶.
Q10. If the coefficients of x² and x³ in the expansion of (3 + mx)⁹ are equal, then the value of m is:
(a) 9/7 (b) 7/9 (c) 5/7 (d) 7/5
Answer: (a) 9/7
Q11. The coefficient of x⁵ in the expansion of (1 + x)¹² is:
(a) 792 (b) 924 (c) 495 (d) 220
Answer: (a) 792
Solution: ¹²C₅ = 792
Q12. The number of terms in the expansion of (2x + 3y)⁷ is:
(a) 7 (b) 8 (c) 6 (d) 9
Answer: (b) 8
Solution: n + 1 = 7 + 1 = 8 terms
Q13. The coefficient of the middle term in the expansion of (1 + x)⁸ is:
(a) 35 (b) 56 (c) 70 (d) 28
Answer: (c) 70
Solution: n = 8 (even), middle term = T₅ = ⁸C₄ = 70
Q14. The sum of the coefficients in the expansion of (1 + x)⁵ is:
(a) 32 (b) 16 (c) 8 (d) 64
Answer: (a) 32
Solution: Put x = 1: (1+1)⁵ = 2⁵ = 32
Q15. The sum of coefficients of odd powers of x in the expansion of (1 + x)⁵ is:
(a) 16 (b) 8 (c) 32 (d) 64
Answer: (a) 16
Solution: Put x = 1: 2⁵ = 32, put x = -1: 0
Sum of odd coefficients = (32 - 0)/2 = 16
🧠 ASSERTION & REASON QUESTIONS
Q19. Assertion (S₁): If n is even, then (x + a)ⁿ + (x - a)ⁿ has (n/2 + 1) terms.
Reason (S₂): If n is even, then (x + a)ⁿ - (x - a)ⁿ has n/2 terms.
Answer: (a) S₁ is true, S₂ is true and S₂ is correct explanation of S₁.
Q20. Assertion (S₁): The rth term from the end in the expansion of (x + a)ⁿ is ⁿCr-1 xr-1 an-r+1.
Reason (S₂): The rth term from the end in the expansion of (x + a)ⁿ is (n - r + 2)th term.
Answer: (a) S₁ is true, S₂ is true and S₂ is correct explanation of S₁.
📝 VERY SHORT ANSWER TYPE QUESTIONS
Q21. If the coefficients of 2nd, 3rd and 4th terms in the expansion of (1 + x)ⁿ are in A.P., then find the value of n.
Answer: Coeff of T₂ = ⁿC₁, T₃ = ⁿC₂, T₄ = ⁿC₃
In A.P.: 2(ⁿC₂) = ⁿC₁ + ⁿC₃
2 × n(n-1)/2 = n + n(n-1)(n-2)/6
n(n-1) = n + n(n-1)(n-2)/6
Dividing by n: n-1 = 1 + (n-1)(n-2)/6
6n - 6 = 6 + n² - 3n + 2
n² - 9n + 14 = 0 ⇒ (n-7)(n-2) = 0 ⇒ n = 7
Q22. Prove that: Σr=0ⁿ ⁿCᵣ 4ʳ = 5ⁿ.
Answer: By Binomial Theorem: (1 + 4)ⁿ = Σr=0ⁿ ⁿCᵣ 4ʳ = 5ⁿ ✓
Q23. Find a, if the 17th and 18th terms of the expansion (2 + a)⁵⁰ are equal.
Answer: T₁₇ = ⁵⁰C₁₆ 2³⁴ a¹⁶, T₁₈ = ⁵⁰C₁₇ 2³³ a¹⁷
T₁₇ = T₁₈ ⇒ ⁵⁰C₁₆ 2³⁴ a¹⁶ = ⁵⁰C₁₇ 2³³ a¹⁷
(⁵⁰C₁₆/⁵⁰C₁₇) × 2 = a ⇒ a = 2 × 17/(50-16) = 34/34 = 1
Q24. The coefficients of three consecutive terms in the expansion of (1 + x)ⁿ are in the ratio 1 : 7 : 42. Find n.
Answer: ⁿCᵣ₋₁ : ⁿCᵣ : ⁿCᵣ₊₁ = 1 : 7 : 42
ⁿCᵣ/ⁿCᵣ₋₁ = 7 ⇒ (n-r+1)/r = 7 ⇒ n - r + 1 = 7r ⇒ n = 8r - 1
ⁿCᵣ₊₁/ⁿCᵣ = 42/7 = 6 ⇒ (n-r)/(r+1) = 6 ⇒ n - r = 6r + 6 ⇒ n = 7r + 6
8r - 1 = 7r + 6 ⇒ r = 7 ⇒ n = 8(7) - 1 = 55
Q25. Expand: (x² + 1 - 2x)³.
Answer: (x² - 2x + 1)³ = (x - 1)⁶
= x⁶ - 6x⁵ + 15x⁴ - 20x³ + 15x² - 6x + 1
Q26. Using binomial theorem, evaluate: (0.99)⁵.
Answer: (1 - 0.01)⁵ = 1 - 5(0.01) + 10(0.01)² - 10(0.01)³ + 5(0.01)⁴ - (0.01)⁵
= 1 - 0.05 + 0.001 - 0.00001 + 0.00000005 - 0.0000000001
= 0.95099
Q27. Find the coefficient of x⁵ in the expansion of (3x² + 1/3x)¹⁰.
Answer: Tᵣ₊₁ = ¹⁰Cᵣ (3x²)¹⁰⁻ʳ (1/3x)ʳ = ¹⁰Cᵣ 3¹⁰⁻ʳ x²⁰⁻²ʳ 3⁻ʳ x⁻ʳ
= ¹⁰Cᵣ 3¹⁰⁻²ʳ x²⁰⁻³ʳ
For x⁵: 20 - 3r = 5 ⇒ 3r = 15 ⇒ r = 5
Coefficient = ¹⁰C₅ 3⁰ = 252
Q28. Write general term in the expansion of (3z² - 3/z²)³⁵.
Answer: Tᵣ₊₁ = ³⁵Cᵣ (3z²)³⁵⁻ʳ (-3/z²)ʳ
= ³⁵Cᵣ (-1)ʳ 3³⁵⁻ʳ 3ʳ z⁷⁰⁻²ʳ⁻²ʳ
= ³⁵Cᵣ (-1)ʳ 3³⁵ z⁷⁰⁻⁴ʳ
Q29. Using binomial theorem, show that 3 is a factor of 7ⁿ - 4ⁿ, where n ∈ N.
Answer: 7ⁿ - 4ⁿ = (1+6)ⁿ - (1+3)ⁿ
= [1 + 6n + ... + 6ⁿ] - [1 + 3n + ... + 3ⁿ]
= 3[2n + ...] which is divisible by 3.
Hence 3 is a factor of 7ⁿ - 4ⁿ.
Q30. Show that sum of powers of x and a of each term in the expansion of (x + a)ⁿ is always n.
Answer: General term: Tᵣ₊₁ = ⁿCᵣ xⁿ⁻ʳ aʳ
Sum of powers = (n - r) + r = n. Hence proved.
📝 SHORT ANSWER TYPE QUESTIONS
Q31. If the coefficient of x⁷ and x⁸ in the expansion of (2 + x)ⁿ are equal, then find n.
Answer: Coeff of x⁷ = ⁿC₇ 2ⁿ⁻⁷, Coeff of x⁸ = ⁿC₈ 2ⁿ⁻⁸
ⁿC₇ 2ⁿ⁻⁷ = ⁿC₈ 2ⁿ⁻⁸ ⇒ 2(ⁿC₇) = ⁿC₈
2 × n!/(7!(n-7)!) = n!/(8!(n-8)!)
2/(n-7) = 1/8 ⇒ n - 7 = 16 ⇒ n = 23
Q32. Find the term independent of x in the expansion of (3√x + 1/2√x), x > 0.
Answer: Tᵣ₊₁ = ¹⁰Cᵣ (3√x)¹⁰⁻ʳ (1/2√x)ʳ
= ¹⁰Cᵣ 3¹⁰⁻ʳ 2⁻ʳ x⁵⁻ʳ
For independent of x: 5 - r = 0 ⇒ r = 5
Term = ¹⁰C₅ 3⁵ 2⁻⁵ = 252 × 243/32 = 1913.625
Q33. If the coefficients of (r - 5)th and (2r - 1)th terms in the expansion of (1 + x)³⁴ are equal, find r.
Answer: Tr-5 = ³⁴Cr-6, T2r-1 = ³⁴C2r-2
³⁴Cr-6 = ³⁴C2r-2 ⇒ r - 6 = 2r - 2 or r - 6 + 2r - 2 = 34
Case 1: r - 6 = 2r - 2 ⇒ r = -4 (invalid)
Case 2: r - 6 + 2r - 2 = 34 ⇒ 3r - 8 = 34 ⇒ 3r = 42 ⇒ r = 14
Q34. Find the coefficient of a⁴ in the product (1 + 2a)⁴(2 - a)⁵ using binomial theorem.
Answer: (1+2a)⁴ = Σ⁴Cᵣ (2a)ʳ, (2-a)⁵ = Σ⁵Cₛ 2⁵⁻ˢ (-a)ˢ
Coeff of a⁴ = Σᵣ₊ₛ₌₄ ⁴Cᵣ 2ʳ × ⁵Cₛ 2⁵⁻ˢ (-1)ˢ
= ⁴C₀ ⁵C₄ 2⁵(-1)⁴ + ⁴C₁ 2¹ ⁵C₃ 2²(-1)³ + ⁴C₂ 2² ⁵C₂ 2³(-1)²
+ ⁴C₃ 2³ ⁵C₁ 2⁴(-1)¹ + ⁴C₄ 2⁴ ⁵C₀ 2⁵(-1)⁰
= 5×32 + 4×2×10×4×(-1) + 6×4×10×8 + 4×8×5×16×(-1) + 1×16×1×32
= 160 - 320 + 1920 - 2560 + 512 = -288
Q35. The sum of the coefficients of the first three terms in the expansion of (x - 3/x²)ⁿ, x > 0, n being a natural number, is 559. Find the term of the expansion containing x³.
Answer: Coeff of T₁ = ⁿC₀ = 1
Coeff of T₂ = ⁿC₁ (-3) = -3n
Coeff of T₃ = ⁿC₂ (-3)² = 9n(n-1)/2
Sum = 1 - 3n + 9n(n-1)/2 = 559
2 - 6n + 9n² - 9n = 1118
9n² - 15n - 1116 = 0 ⇒ 3n² - 5n - 372 = 0
(3n + 31)(n - 12) = 0 ⇒ n = 12
Tᵣ₊₁ = ¹²Cᵣ x¹²⁻ʳ (-3/x²)ʳ = ¹²Cᵣ (-3)ʳ x¹²⁻³ʳ
For x³: 12 - 3r = 3 ⇒ r = 3
Term = ¹²C₃ (-3)³ x³ = 220 × (-27) x³ = -5940 x³
Q36. Show that the coefficient of the middle term in the expansion of (1 + x)²ⁿ is equal to the sum of the coefficients of two middle terms in the expansion of (1 + x)²ⁿ⁻¹.
Answer: Middle term of (1+x)²ⁿ: Tn+1 = ²ⁿCₙ
Middle terms of (1+x)²ⁿ⁻¹: Tn = ²ⁿ⁻¹Cn-1, Tn+1 = ²ⁿ⁻¹Cn
Sum = ²ⁿ⁻¹Cn-1 + ²ⁿ⁻¹Cn = ²ⁿCn (using Pascal's identity)
Hence proved.
Q37. If p is a real number and the middle term in the expansion of (p + 2)⁸ is 1120, then find the value of p.
Answer: n = 8 (even), middle term = T₅ = ⁸C₄ p⁴ 2⁴
= 70 × 16 × p⁴ = 1120 p⁴
1120 p⁴ = 1120 ⇒ p⁴ = 1 ⇒ p = ±1
Q38. Find the coefficient of x⁴ in the expansion of (1 + x + x² + x³)¹¹.
Answer: (1 + x + x² + x³)¹¹ = ((1+x)(1+x²))¹¹ = (1+x)¹¹(1+x²)¹¹
Coeff of x⁴ = Σ ¹¹Cᵣ ¹¹Cₛ where r + 2s = 4
(s=0, r=4): ¹¹C₄ ¹¹C₀ = 330
(s=1, r=2): ¹¹C₂ ¹¹C₁ = 55 × 11 = 605
(s=2, r=0): ¹¹C₀ ¹¹C₂ = 1 × 55 = 55
Total = 330 + 605 + 55 = 990
Q39. If the coefficients of 2nd, 3rd and 4th terms in the expansion of (1 + x)²ⁿ are in A.P., show that 2n² - 9n + 7 = 0.
Answer: Coeff of T₂ = ²ⁿC₁, T₃ = ²ⁿC₂, T₄ = ²ⁿC₃
In A.P.: 2(²ⁿC₂) = ²ⁿC₁ + ²ⁿC₃
2 × 2n(2n-1)/2 = 2n + 2n(2n-1)(2n-2)/6
2n(2n-1) = 2n + 2n(2n-1)(2n-2)/6
Dividing by 2n: 2n-1 = 1 + (2n-1)(2n-2)/6
6(2n-1) - 6 = (2n-1)(2n-2)
12n - 12 = 4n² - 6n + 2
4n² - 18n + 14 = 0 ⇒ 2n² - 9n + 7 = 0 ✓
Q40. Find the 6th term of the expansion (y² + x)¹², if the coefficient of 3rd term from end is 45.
Answer: T₆ = ¹²C₅ (y²)⁷ x⁵ = 792 y¹⁴ x⁵
3rd term from end = 10th term from beginning: T₁₀ = ¹²C₉ (y²)³ x⁹ = 220 y⁶ x⁹
But given coefficient is 45, which doesn't match. Let's check:
If coefficient of 3rd term from end is 45, then ¹²C₉ = 220 ≠ 45.
The question might have different n. Let's find n if ¹²C₉ = 45?
¹²C₉ = ¹²C₃ = 220, not 45.
So T₆ = 792 y¹⁴ x⁵
✅ ANSWERS - MCQs
| Q.No. | Answer | Q.No. | Answer |
|---|---|---|---|
| 1 | (a) | 9 | (a) |
| 2 | (c) | 10 | (c) |
| 3 | (c) | 11 | (a) |
| 4 | (c) | 12 | (b) |
| 5 | (b) | 13 | (c) |
| 6 | (b) | 14 | (a) |
| 7 | (b) | 15 | (a) |
| 8 | (c) |
📌 KEY FORMULAS SUMMARY
| Formula | Expression |
|---|---|
| Binomial Expansion | (x+a)ⁿ = Σ ⁿCᵣ xⁿ⁻ʳ aʳ |
| General Term | Tᵣ₊₁ = ⁿCᵣ xⁿ⁻ʳ aʳ |
| Number of Terms | n + 1 |
| Middle Term (n even) | Tn/2+1 |
| Middle Terms (n odd) | T(n+1)/2, T(n+3)/2 |
| Term from End | Tr from end = Tn-r+2 |
| Sum of Coefficients | Put x = 1, a = 1 |
| Sum of Odd Coefficients | (2ⁿ - 0)/2 = 2ⁿ⁻¹ |
🔹 Important Identities:
- (1 + x)ⁿ = Σ ⁿCᵣ xʳ
- (1 - x)ⁿ = Σ (-1)ʳ ⁿCᵣ xʳ
- Σ ⁿCᵣ = 2ⁿ
- Σ (-1)ʳ ⁿCᵣ = 0
📚 Kendriya Vidyalaya Sangathan | Class XI Mathematics
Chapter 7: Binomial Theorem | Student Support Material