Binomial Theorem — Class 11 Mathematics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 11 Mathematics chapter "Binomial Theorem" — 8 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 11 Mathematics chapter "Binomial Theorem" — 8 important questions with detailed answers for CBSE board exa…
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NCERT Solutions for Class 11 Mathematics — Binomial Theorem. Step-by-step answers to all 8 textbook questions from this chapter, written for CBSE board preparation and free to use.
Binomial Theorem — All 8 Questions Solved
Q1. Find the binomial expansion of (2x + 3)⁴.
Step 1: Identify the binomial theorem formula.
(a + b)ⁿ = Σ(k=0 to n) [nCk × aⁿ⁻ᵏ × bᵏ]
= nC0 a^n + nC1 a^(n-1) b + nC2 a^(n-2) b² + ... + nCn b^n
Step 2: Identify parameters.
a = 2x, b = 3, n = 4
Step 3: Calculate binomial coefficients.
4C0 = 1
4C1 = 4
4C2 = 6
4C3 = 4
4C4 = 1
Step 4: Expand term by term.
Term 1: 4C0 (2x)⁴ (3)⁰ = 1 × 16x⁴ × 1 = 16x⁴
Term 2: 4C1 (2x)³ (3)¹ = 4 × 8x³ × 3 = 96x³
Term 3: 4C2 (2x)² (3)² = 6 × 4x² × 9 = 216x²
Term 4: 4C3 (2x)¹ (3)³ = 4 × 2x × 27 = 216x
Term 5: 4C4 (2x)⁰ (3)⁴ = 1 × 1 × 81 = 81
Step 5: Write the complete expansion.
(2x + 3)⁴ = 16x⁴ + 96x³ + 216x² + 216x + 81
Final Answer: (2x + 3)⁴ = 16x⁴ + 96x³ + 216x² + 216x + 81
Q2. Find the 5th term in the expansion of (x - 2y)⁸.
Step 1: Recall the general term formula in binomial expansion.
Tr+1 = nCr × a^(n-r) × b^r
where the (r+1)th term is given, with r starting from 0.
Step 2: Identify parameters for the 5th term.
We need T5, which means r + 1 = 5, so r = 4.
a = x, b = -2y, n = 8
Step 3: Calculate the binomial coefficient.
8C4 = 8! / (4! × 4!)
= (8 × 7 × 6 × 5) / (4 × 3 × 2 × 1)
= 1680 / 24
= 70
Step 4: Calculate the powers.
(x)^(8-4) = x⁴
(-2y)⁴ = 16y⁴
Step 5: Find the 5th term.
T5 = 8C4 × x⁴ × (-2y)⁴
= 70 × x⁴ × 16y⁴
= 1120x⁴y⁴
Step 6: Verify the sign.
(-2y)⁴ = (-2)⁴ × y⁴ = 16y⁴ (positive, since the power is even)
Final Answer: The 5th term is 1120x⁴y⁴
Q3. Find the middle term(s) in the expansion of (p + q)¹⁰.
Step 1: Determine the total number of terms.
For (a + b)ⁿ, the total number of terms = n + 1
For (p + q)¹⁰, total terms = 10 + 1 = 11 terms
Step 2: Find the middle term(s).
Since there are 11 terms (odd), there is exactly one middle term.
The middle term is the (11 + 1)/2 = 6th term.
Step 3: Identify r for the 6th term.
For T(r+1), we need r + 1 = 6, so r = 5.
Step 4: Apply the general term formula.
T6 = 10C5 × p^(10-5) × q⁵
= 10C5 × p⁵ × q⁵
Step 5: Calculate the binomial coefficient.
10C5 = 10! / (5! × 5!)
= (10 × 9 × 8 × 7 × 6) / (5 × 4 × 3 × 2 × 1)
= 30,240 / 120
= 252
Step 6: Write the middle term.
T6 = 252p⁵q⁵
Step 7: Verify.
The middle term should have equal powers of p and q.
252p⁵q⁵ has power 5 for both, which confirms it's the middle term.
Final Answer: The middle term is 252p⁵q⁵
Q4. Find the coefficient of x⁷ in the expansion of (2x + 3)¹⁰.
Step 1: Use the general term of binomial expansion.
Tr+1 = nCr × a^(n-r) × b^r
For (2x + 3)¹⁰:
a = 2x, b = 3, n = 10
Step 2: Write the general term.
Tr+1 = 10Cr × (2x)^(10-r) × 3^r
= 10Cr × 2^(10-r) × x^(10-r) × 3^r
Step 3: Find r for which the power of x is 7.
10 - r = 7
r = 3
Step 4: Calculate the coefficient of x⁷.
T(3+1) = T4 = 10C3 × 2^(10-3) × 3³
= 10C3 × 2⁷ × 3³
Step 5: Calculate each component.
10C3 = 10! / (3! × 7!)
= (10 × 9 × 8) / (3 × 2 × 1)
= 720 / 6
= 120
2⁷ = 128
3³ = 27
Step 6: Find the coefficient.
Coefficient of x⁷ = 120 × 128 × 27
= 120 × 128 × 27
= 15,360 × 27
= 414,720
Final Answer: The coefficient of x⁷ is 414,720
Q5. Prove that the sum of binomial coefficients in the expansion of (x + y)ⁿ is 2ⁿ.
Step 1: Write the binomial expansion.
(x + y)ⁿ = nC0 x^n + nC1 x^(n-1) y + nC2 x^(n-2) y² + ... + nCn y^n
= Σ(k=0 to n) [nCk × x^(n-k) × y^k]
Step 2: State the sum of binomial coefficients.
Sum = nC0 + nC1 + nC2 + ... + nCn
Step 3: Substitute x = 1 and y = 1 in the binomial expansion.
(1 + 1)ⁿ = nC0 (1)ⁿ + nC1 (1)^(n-1) (1) + nC2 (1)^(n-2) (1)² + ... + nCn (1)ⁿ
2ⁿ = nC0 + nC1 + nC2 + ... + nCn
Step 4: Conclusion.
Therefore, the sum of binomial coefficients is 2ⁿ.
Step 5: Verify with an example.
For n = 3:
(x + y)³ = 3C0 + 3C1 + 3C2 + 3C3
= 1 + 3 + 3 + 1
= 8
= 2³ ✓
Final Answer: nC0 + nC1 + nC2 + ... + nCn = 2ⁿ is proved.
Q6. Find the coefficient of x in the expansion of (3x² - 1/x)⁵.
Step 1: Write the general term.
Tr+1 = 5Cr × (3x²)^(5-r) × (-1/x)^r
= 5Cr × 3^(5-r) × x^(2(5-r)) × (-1)^r × x^(-r)
= 5Cr × 3^(5-r) × (-1)^r × x^(10-2r-r)
= 5Cr × 3^(5-r) × (-1)^r × x^(10-3r)
Step 2: Find r such that the power of x is 1.
10 - 3r = 1
3r = 9
r = 3
Step 3: Calculate the coefficient of x.
T(3+1) = T4 = 5C3 × 3^(5-3) × (-1)³
= 5C3 × 3² × (-1)
Step 4: Calculate each component.
5C3 = 5! / (3! × 2!)
= (5 × 4) / (2 × 1)
= 10
3² = 9
(-1)³ = -1
Step 5: Find the coefficient.
Coefficient of x = 10 × 9 × (-1)
= -90
Final Answer: The coefficient of x is -90
Q7. If (1 + x)ⁿ = 1 + 6x + 15x² + 20x³ + ..., find the value of n.
Step 1: Write the general term of (1 + x)ⁿ.
Tr+1 = nCr × x^r
Step 2: Compare coefficients.
From the given expansion:
- Coefficient of x: nC1 = 6
- Coefficient of x²: nC2 = 15
- Coefficient of x³: nC3 = 20
Step 3: Use nC1 = 6.
nC1 = n = 6
Step 4: Verify with nC2 = 15.
6C2 = 6! / (2! × 4!)
= (6 × 5) / 2
= 15 ✓
Step 5: Verify with nC3 = 20.
6C3 = 6! / (3! × 3!)
= (6 × 5 × 4) / (3 × 2 × 1)
= 120 / 6
= 20 ✓
Step 6: Conclusion.
All three conditions are satisfied for n = 6.
Final Answer: n = 6
Q8. Using the binomial theorem, find the approximate value of (0.99)⁵.
Step 1: Express 0.99 in a suitable form.
0.99 = 1 - 0.01
Step 2: Apply the binomial theorem.
(1 - 0.01)⁵ = Σ(k=0 to 5) [5Ck × 1^(5-k) × (-0.01)^k]
= 5C0 - 5C1(0.01) + 5C2(0.01)² - 5C3(0.01)³ + ...
Step 3: Calculate the terms.
5C0 = 1
5C1 = 5
5C2 = 10
5C3 = 10
Step 4: Expand with increasing powers of 0.01.
Term 1: 5C0 × 1 = 1
Term 2: -5C1 × 0.01 = -5 × 0.01 = -0.05
Term 3: 5C2 × (0.01)² = 10 × 0.0001 = 0.001
Term 4: -5C3 × (0.01)³ = -10 × 0.000001 = -0.00001 (negligible)
Step 5: Sum the significant terms.
(0.99)⁵ ≈ 1 - 0.05 + 0.001
= 0.951
Step 6: Note about precision.
Terms beyond the third become very small and can be ignored for a good approximation.
Final Answer: (0.99)⁵ ≈ 0.951
Binomial Theorem — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. How many terms are there in the expansion of (2x + 3y)^10?
- A. 10
- B. 11
- C. 12
- D. 9
Answer: B. 11 — The expansion of (a+b)^n contains (n+1) terms. Here n=10, so there are 10+1 = 11 terms.
MCQ 2. What is the general term in the expansion of (a+b)^n?
- A. nCr * a^(n-r) * b^r
- B. nCr * a^r * b^(n-r)
- C. nPr * a^(n-r) * b^r
- D. nPr * a^r * b^(n-r)
Answer: A. nCr * a^(n-r) * b^r — The general term (r+1)th term in the binomial expansion of (a+b)^n is given by T_(r+1) = nCr * a^(n-r) * b^r.
MCQ 3. Find the 4th term in the expansion of (x + 2)^5.
- A. 40x^2
- B. 80x^2
- C. 40x^3
- D. 80x^3
Answer: B. 80x^2 — For the 4th term, r=3. T_4 = 5C3 * x^(5-3) * (2)^3 = 10 * x^2 * 8 = 80x^2.
MCQ 4. What is the coefficient of x^3 in the expansion of (1 + 2x)^5?
- A. 40
- B. 80
- C. 20
- D. 10
Answer: B. 80 — The general term is T_(r+1) = 5Cr * 1^(5-r) * (2x)^r = 5Cr * 2^r * x^r. For x^3, r=3. Coefficient = 5C3 * 2^3 = 10 * 8 = 80.
MCQ 5. Find the term independent of x in the expansion of (x^2 + 1/x)^9.
- A. 84
- B. 126
- C. 36
- D. 9
Answer: A. 84 — The general term is T_(r+1) = 9Cr * (x^2)^(9-r) * (1/x)^r = 9Cr * x^(18-3r). For term independent of x, 18-3r = 0, so r=6. The term is 9C6 = 9C3 = 84.
MCQ 6. What is the sum of the binomial coefficients in the expansion of (1+x)^n?
- A. 2n
- B. n^2
- C. 2^n
- D. n+1
Answer: C. 2^n — The sum of binomial coefficients in the expansion of (1+x)^n is obtained by substituting x=1, which gives (1+1)^n = 2^n.
MCQ 7. Find the middle term in the expansion of (x/2 + 2/x)^8.
- A. 70
- B. 70x^4
- C. 70/x^4
- D. 56
Answer: A. 70 — For n=8 (even), the middle term is the (n/2 + 1)th term, which is the 5th term. T_5 = 8C4 * (x/2)^4 * (2/x)^4 = 8C4 = 70.
MCQ 8. How many middle terms are there in the expansion of (a+b)^9?
- A. 1
- B. 2
- C. 3
- D. 0
Answer: B. 2 — When n is odd, there are two middle terms, given by ((n+1)/2)th and ((n+1)/2 + 1)th terms. For n=9, these are the 5th and 6th terms.
MCQ 9. If the coefficients of the (2r+1)th and (r+2)th terms in the expansion of (1+x)^43 are equal, then the value of r is:
- A. 12
- B. 13
- C. 14
- D. 15
Answer: C. 14 — The coefficient of the (2r+1)th term is 43C(2r). The coefficient of the (r+2)th term is 43C(r+1). If they are equal, then 2r + (r+1) = 43, which gives 3r+1=43, so 3r=42 and r=14.
MCQ 10. Which of the following is true for binomial coefficients?
- A. nCr = nC(r-1)
- B. nCr = nC(n-r)
- C. nCr = (n-1)Cr
- D. nCr = nC(r+1)
Answer: B. nCr = nC(n-r) — This is a fundamental property of binomial coefficients, stating that the number of ways to choose r items from n is the same as choosing (n-r) items to leave behind.
MCQ 11. The 4th term in the expansion of (x - 1/x)^8 is:
- A. 56x^2
- B. -56x^2
- C. 56x^4
- D. -56x^4
Answer: B. -56x^2 — For the 4th term, r=3. T_4 = 8C3 * x^(8-3) * (-1/x)^3 = 56 * x^5 * (-1/x^3) = -56x^2.
MCQ 12. The coefficient of x^(-2) in the expansion of (x - 1/x^2)^10 is:
- A. 210
- B. -210
- C. 45
- D. -45
Answer: A. 210 — The general term is T_(r+1) = 10Cr * x^(10-r) * (-1/x^2)^r = 10Cr * (-1)^r * x^(10-3r). For x^(-2), 10-3r = -2, so r=4. Coefficient = 10C4 * (-1)^4 = 210.
MCQ 13. The third term in the expansion of (2 + x)^5 is:
- A. 80x^2
- B. 40x^3
- C. 20x^2
- D. 10x^3
Answer: A. 80x^2 — For the third term, r=2. T_3 = 5C2 * 2^(5-2) * x^2 = 10 * 8 * x^2 = 80x^2.
MCQ 14. The number of irrational terms in the expansion of (√3 + ⁵√2)^10 is:
- A. 8
- B. 9
- C. 10
- D. 11
Answer: B. 9 — The general term is 10Cr * 3^((10-r)/2) * 2^(r/5). For a rational term, (10-r)/2 and r/5 must be integers. This occurs for r=0 and r=10. Since there are 11 total terms, 11-2 = 9 terms are irrational.
MCQ 15. The sum of the odd binomial coefficients in the expansion of (1+x)^7 is:
- A. 64
- B. 128
- C. 256
- D. 32
Answer: A. 64 — The sum of the odd binomial coefficients in the expansion of (1+x)^n is 2^(n-1). For n=7, the sum is 2^(7-1) = 2^6 = 64.
MCQ 16. C(n,r) + C(n,r-1) is equal to:
- A. C(n+1,r)
- B. C(n+1,r-1)
- C. C(n,r+1)
- D. C(n-1,r)
Answer: A. C(n+1,r) — This is Pascal's rule for binomial coefficients, which states that nCr + nC(r-1) = (n+1)Cr.
MCQ 17. The number of terms in the expansion of (a+b+c)^n is:
- A. n+1
- B. (n+1)(n+2)/2
- C. n^2+1
- D. (n+1)^2
Answer: B. (n+1)(n+2)/2 — For a multinomial expansion like (a+b+c)^n, the number of terms is C(n+k-1, k-1) where k is the number of terms inside the parenthesis. Here k=3, so C(n+2, 2) = (n+2)(n+1)/2.
MCQ 18. The expansion of (1-x)^n is given by:
- A. Σ nCr * x^r
- B. Σ nCr * (-x)^r
- C. Σ (-1)^r * nCr * x^r
- D. Σ (-1)^r * nCr * x^(n-r)
Answer: C. Σ (-1)^r * nCr * x^r — The expansion of (1-x)^n is given by the sum from r=0 to n of nCr * (1)^(n-r) * (-x)^r = Σ nCr * (-1)^r * x^r.
MCQ 19. The remainder when 9^10 is divided by 8 is:
- A. 0
- B. 1
- C. 2
- D. 7
Answer: B. 1 — Write 9^10 as (1+8)^10. Using the binomial theorem, this expands to 1 + (a multiple of 8). Thus, the remainder is 1.
MCQ 20. What is the highest power of x in the expansion of (x^2 + 1/x)^10?
- A. x^10
- B. x^20
- C. x^15
- D. x^12
Answer: B. x^20 — The highest power of x occurs in the first term (r=0) where the (1/x) part is not raised to any power, i.e., 10C0 * (x^2)^10 = x^20.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.