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.
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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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Key Questions Covered:
- Find the binomial expansion of (2x + 3)⁴.
- Find the 5th term in the expansion of (x - 2y)⁸.
- Find the middle term(s) in the expansion of (p + q)¹⁰.
- Find the coefficient of x⁷ in the expansion of (2x + 3)¹⁰.
- Prove that the sum of binomial coefficients in the expansion of (x + y)ⁿ is 2ⁿ.
- Find the coefficient of x in the expansion of (3x² - 1/x)⁵.
- + 2 more questions in the full chapter
Solutions Summary:
| Question | Status |
|---|---|
| Find the binomial expansion of (2x + 3)⁴. | ✓ Solved |
| Find the 5th term in the expansion of (x - 2y)⁸. | ✓ Solved |
| Find the middle term(s) in the expansion of (p + q)¹⁰. | ✓ Solved |
| Find the coefficient of x⁷ in the expansion of (2x + 3)¹⁰. | ✓ Solved |
| Prove that the sum of binomial coefficients in the expans… | ✓ Solved |
| Find the coefficient of x in the expansion of (3x² - 1/x)⁵. | ✓ Solved |
Showing 6 of 8 questions
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 ...
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...
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 =...
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
St...
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 examp...
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...
Showing 6 of 8 questions. Visit the full page for complete solutions.
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