Important Questions Class 11 Maths Chapter 7 Binomial Theorem

Binomial Theorem gives the expansion of (a + b)ⁿ for a positive integral value of n. Its coefficients come from combinations and can also be read from Pascal’s triangle.

Large powers become manageable when the coefficient pattern is known. Important Questions Class 11 Maths Chapter 7 help students practise Pascal’s triangle, binomial coefficients, expansion of binomials, special cases, approximation, divisibility, and coefficient-based questions. The 2026 NCERT chapter Binomial Theorem studies expansions for positive integral indices only, with applications such as evaluating (98)⁵, expanding expressions, and proving divisibility statements.

Key Takeaways

• Binomial Theorem: (a + b)ⁿ = ⁿC₀aⁿ + ⁿC₁aⁿ⁻¹b + ⁿC₂aⁿ⁻²b² + ... + ⁿCₙbⁿ.
• Number Of Terms: The expansion of (a + b)ⁿ has n + 1 terms.
• Binomial Coefficient: ⁿCᵣ = n!/[r!(n − r)!] gives the coefficient of the (r + 1)th term.
• Negative Sign Pattern: In (x − y)ⁿ, signs alternate as + − + − ....

Important Questions Class 11 Maths Chapter 7 Structure 2026

Binomial Theorem Class 11 Chapter Overview

Binomial Theorem Class 11 gives a direct method to expand powers of two-term expressions. It avoids repeated multiplication for powers such as (101)⁴ and (98)⁵.

The chapter begins with known expansions like (a + b)² and (a + b)³. It then connects the coefficient pattern with Pascal’s triangle and combinations.

Q1. What Is Binomial Theorem Class 11?

Binomial Theorem Class 11 states the expansion of (a + b)ⁿ for a positive integer n.

It writes the expansion using binomial coefficients. The theorem applies to positive integral indices in the 2026 NCERT chapter.

Formula:
(a + b)ⁿ = ⁿC₀aⁿ + ⁿC₁aⁿ⁻¹b + ⁿC₂aⁿ⁻²b² + ... + ⁿCₙbⁿ

Q2. Why Are Important Questions Class 11 Maths Chapter 7 Useful For Exams?

Important Questions Class 11 Maths Chapter 7 are useful because expansion, coefficient, and divisibility questions appear often.

Students must recognise the correct form before expanding. Sign errors are common in (x − y)ⁿ and (1 − x)ⁿ.

Example: (1 − x)ⁿ = ⁿC₀ − ⁿC₁x + ⁿC₂x² − ... + (−1)ⁿⁿCₙxⁿ.

Q3. What Does Positive Integral Index Mean In Binomial Theorem?

A positive integral index means n is a positive integer such as 1, 2, 3, or 4.

The 2026 NCERT Chapter 7 treats Binomial Theorem for positive integral indices only. It does not develop the infinite binomial series for rational indices.

Example: (a + b)⁵ is included, while (1 + x)¹/² is outside this chapter.

Pascal Triangle Class 11 And Binomial Coefficients

Pascal’s triangle gives binomial coefficients row by row. The row for index n gives coefficients of (a + b)ⁿ.

Each row begins and ends with 1. Every inside number comes from adding the two numbers above it.

Q4. What Is Pascal Triangle Class 11?

Pascal’s triangle is a triangular arrangement of binomial coefficients.

It starts with 1 at the top. The row for index 5 is 1, 5, 10, 10, 5, 1.

Example: (a + b)⁵ uses coefficients 1, 5, 10, 10, 5, 1.

Q5. What Are Binomial Coefficients Class 11?

Binomial coefficients are the numbers ⁿC₀, ⁿC₁, ⁿC₂, ..., ⁿCₙ in the expansion of (a + b)ⁿ.

They multiply the terms in the expansion. They can be found using combinations.

Formula:
ⁿCᵣ = n!/[r!(n − r)!]

Q6. Write The Coefficients In The Expansion Of (a + b)⁷.

The coefficients are 1, 7, 21, 35, 35, 21, 7, 1.

1. Given expression:
   (a + b)⁷
2. Required row:
   Index 7 in Pascal’s triangle
3. Coefficients:
   ⁷C₀, ⁷C₁, ⁷C₂, ⁷C₃, ⁷C₄, ⁷C₅, ⁷C₆, ⁷C₇
4. Values:
   1, 7, 21, 35, 35, 21, 7, 1

Final Result: 1, 7, 21, 35, 35, 21, 7, 1

Q7. Find ⁸C₃.

The value of ⁸C₃ is 56.

1. Formula Used:
   ⁿCᵣ = n!/[r!(n − r)!]
2. Substitute:
   ⁸C₃ = 8!/[3!5!]
3. Cancel:
   ⁸C₃ = (8 × 7 × 6)/(3 × 2 × 1)
4. Calculate:
   ⁸C₃ = 56

Final Result: ⁸C₃ = 56

Q8. Why Are Coefficients Symmetric In Binomial Expansion?

Coefficients are symmetric because ⁿCᵣ = ⁿCₙ₋ᵣ.

The coefficient near the beginning matches the coefficient at the same distance from the end. This symmetry appears in Pascal’s triangle.

Example: In row 7, ⁷C₂ = ⁷C₅ = 21.

Binomial Theorem Formula Class 11 Important Questions

The formula keeps the power of the first term decreasing and the power of the second term increasing. The sum of powers stays equal to n.

This pattern helps students write expansions without missing any term.

Q9. State The Binomial Theorem Formula Class 11.

The Binomial Theorem formula is (a + b)ⁿ = ⁿC₀aⁿ + ⁿC₁aⁿ⁻¹b + ⁿC₂aⁿ⁻²b² + ... + ⁿCₙbⁿ.

Here, n is a positive integer. The expansion has n + 1 terms.

Final Result: Binomial Theorem formula stated

Q10. What Is The General Term In Binomial Expansion?

The general term in binomial expansion is Tᵣ₊₁ = ⁿCᵣ aⁿ⁻ʳbʳ.

It gives the (r + 1)th term in the expansion of (a + b)ⁿ. The value of r starts from 0.

Example: For the third term, use r = 2.

Q11. How Many Terms Are There In (a + b)⁹?

There are 10 terms in the expansion of (a + b)⁹.

1. Given index:
   n = 9
2. Rule Used:
   Number of terms = n + 1
3. Calculation:
   9 + 1 = 10

Final Result: 10 terms

Q12. What Is The Sum Of Powers In Each Term Of (a + b)ⁿ?

The sum of powers in each term is always n.

In the term ⁿCᵣaⁿ⁻ʳbʳ, the powers are n − r and r. Their sum is n.

Calculation:
(n − r) + r = n

Final Result: Sum of powers = n

Expansion Of Binomial Class 11 Questions With Answers

Expansion questions need clean substitution into the theorem. Students should first identify a, b, and n.

When the second term is negative, keep the sign inside the bracket until the powers are applied.

Q13. Expand (x + 2)⁶ Using Binomial Theorem.

The expansion is x⁶ + 12x⁵ + 60x⁴ + 160x³ + 240x² + 192x + 64.

1. Given Data:
   a = x, b = 2, n = 6
2. Formula Used:
   (a + b)⁶ = ⁶C₀a⁶ + ⁶C₁a⁵b + ⁶C₂a⁴b² + ... + ⁶C₆b⁶
3. Substitute:
   (x + 2)⁶ = ⁶C₀x⁶ + ⁶C₁x⁵(2) + ⁶C₂x⁴(2²) + ⁶C₃x³(2³) + ⁶C₄x²(2⁴) + ⁶C₅x(2⁵) + ⁶C₆(2⁶)
4. Calculate:
   = x⁶ + 12x⁵ + 60x⁴ + 160x³ + 240x² + 192x + 64

Final Result: x⁶ + 12x⁵ + 60x⁴ + 160x³ + 240x² + 192x + 64

Q14. Expand (2x + 3y)⁵ Using Pascal’s Triangle.

The expansion is 32x⁵ + 240x⁴y + 720x³y² + 1080x²y³ + 810xy⁴ + 243y⁵.

1. Coefficients for index 5:
   1, 5, 10, 10, 5, 1
2. Write expansion:
   (2x + 3y)⁵ = (2x)⁵ + 5(2x)⁴(3y) + 10(2x)³(3y)² + 10(2x)²(3y)³ + 5(2x)(3y)⁴ + (3y)⁵
3. Calculate terms:
   32x⁵ + 240x⁴y + 720x³y² + 1080x²y³ + 810xy⁴ + 243y⁵

Final Result: 32x⁵ + 240x⁴y + 720x³y² + 1080x²y³ + 810xy⁴ + 243y⁵

Q15. Expand (x − 2y)⁵.

The expansion is x⁵ − 10x⁴y + 40x³y² − 80x²y³ + 80xy⁴ − 32y⁵.

1. Given Data:
   a = x, b = −2y, n = 5
2. Formula Used:
   (a + b)⁵ = a⁵ + 5a⁴b + 10a³b² + 10a²b³ + 5ab⁴ + b⁵
3. Substitute:
   (x − 2y)⁵ = x⁵ + 5x⁴(−2y) + 10x³(−2y)² + 10x²(−2y)³ + 5x(−2y)⁴ + (−2y)⁵
4. Simplify:
   x⁵ − 10x⁴y + 40x³y² − 80x²y³ + 80xy⁴ − 32y⁵

Final Result: x⁵ − 10x⁴y + 40x³y² − 80x²y³ + 80xy⁴ − 32y⁵

Q16. Expand (1 − 2x)⁵.

The expansion is 1 − 10x + 40x² − 80x³ + 80x⁴ − 32x⁵.

1. Given Data:
   a = 1, b = −2x, n = 5
2. Use formula:
   (1 − 2x)⁵ = 1 + 5(−2x) + 10(−2x)² + 10(−2x)³ + 5(−2x)⁴ + (−2x)⁵
3. Calculate:
   = 1 − 10x + 40x² − 80x³ + 80x⁴ − 32x⁵

Final Result: 1 − 10x + 40x² − 80x³ + 80x⁴ − 32x⁵

Q17. Expand (2x − 3)⁶.

The expansion is 64x⁶ − 576x⁵ + 2160x⁴ − 4320x³ + 4860x² − 2916x + 729.

1. Given Data:
   a = 2x, b = −3, n = 6
2. Formula Used:
   (a + b)⁶ = a⁶ + 6a⁵b + 15a⁴b² + 20a³b³ + 15a²b⁴ + 6ab⁵ + b⁶
3. Substitute and simplify:
   (2x − 3)⁶ = 64x⁶ − 576x⁵ + 2160x⁴ − 4320x³ + 4860x² − 2916x + 729

Final Result: 64x⁶ − 576x⁵ + 2160x⁴ − 4320x³ + 4860x² − 2916x + 729

Special Cases Of Binomial Theorem Class 11

Special cases help in fast expansion and proof questions. NCERT lists (x − y)ⁿ, (1 + x)ⁿ, and (1 − x)ⁿ as key forms.

These forms also explain coefficient sums and alternating coefficient sums.

Q18. Write The Expansion Of (1 + x)ⁿ.

The expansion is (1 + x)ⁿ = ⁿC₀ + ⁿC₁x + ⁿC₂x² + ... + ⁿCₙxⁿ.

It follows from Binomial Theorem by taking a = 1 and b = x. Each term contains a rising power of x.

Final Result: (1 + x)ⁿ expansion stated

Q19. Write The Expansion Of (1 − x)ⁿ.

The expansion is (1 − x)ⁿ = ⁿC₀ − ⁿC₁x + ⁿC₂x² − ... + (−1)ⁿⁿCₙxⁿ.

It follows by taking b = −x. Odd powers of x carry negative signs.

Final Result: (1 − x)ⁿ expansion stated

Q20. Prove That ⁿC₀ + ⁿC₁ + ⁿC₂ + ... + ⁿCₙ = 2ⁿ.

The sum of all binomial coefficients is 2ⁿ.

1. Use special case:
   (1 + x)ⁿ = ⁿC₀ + ⁿC₁x + ⁿC₂x² + ... + ⁿCₙxⁿ
2. Put x = 1:
   (1 + 1)ⁿ = ⁿC₀ + ⁿC₁ + ⁿC₂ + ... + ⁿCₙ
3. Simplify:
   2ⁿ = ⁿC₀ + ⁿC₁ + ⁿC₂ + ... + ⁿCₙ

Final Result: ⁿC₀ + ⁿC₁ + ⁿC₂ + ... + ⁿCₙ = 2ⁿ

Q21. Prove That ⁿC₀ − ⁿC₁ + ⁿC₂ − ... + (−1)ⁿⁿCₙ = 0.

The alternating sum of binomial coefficients is 0.

1. Use special case:
   (1 − x)ⁿ = ⁿC₀ − ⁿC₁x + ⁿC₂x² − ... + (−1)ⁿⁿCₙxⁿ
2. Put x = 1:
   (1 − 1)ⁿ = ⁿC₀ − ⁿC₁ + ⁿC₂ − ... + (−1)ⁿⁿCₙ
3. Simplify:
   0 = ⁿC₀ − ⁿC₁ + ⁿC₂ − ... + (−1)ⁿⁿCₙ

Final Result: Alternating sum = 0

General Term In Binomial Expansion Questions

The general term helps find one term without writing the full expansion. It is written as the (r + 1)th term.

Students must remember that the first term comes from r = 0.

Q22. Find The General Term In (2x + 3y)⁸.

The general term is Tᵣ₊₁ = ⁸Cᵣ(2x)⁸⁻ʳ(3y)ʳ.

1. Given expression:
   (2x + 3y)⁸
2. Formula Used:
   Tᵣ₊₁ = ⁿCᵣaⁿ⁻ʳbʳ
3. Substitute:
   n = 8, a = 2x, b = 3y
4. General term:
   Tᵣ₊₁ = ⁸Cᵣ(2x)⁸⁻ʳ(3y)ʳ

Final Result: Tᵣ₊₁ = ⁸Cᵣ(2x)⁸⁻ʳ(3y)ʳ

Q23. Find The 4th Term In (x + 2)⁶.

The 4th term is 160x³.

1. For 4th term:
   r + 1 = 4, so r = 3
2. Formula Used:
   Tᵣ₊₁ = ⁿCᵣaⁿ⁻ʳbʳ
3. Substitute values:
   T₄ = ⁶C₃x⁶⁻³(2)³
4. Calculate:
   T₄ = 20 × x³ × 8
5. Simplify:
   T₄ = 160x³

Final Result: 160x³

Q24. Find The Term Containing x³ In (x + 2)⁶.

The term containing x³ is 160x³.

1. General term:
   Tᵣ₊₁ = ⁶Cᵣx⁶⁻ʳ(2)ʳ
2. For x³:
   6 − r = 3
3. Solve:
   r = 3
4. Substitute:
   T₄ = ⁶C₃x³(2)³
5. Calculate:
   T₄ = 20 × 8 × x³ = 160x³

Final Result: 160x³

Q25. Find The Coefficient Of x⁴ In (x + 2)⁶.

The coefficient of x⁴ is 60.

1. General term:
   Tᵣ₊₁ = ⁶Cᵣx⁶⁻ʳ(2)ʳ
2. For x⁴:
   6 − r = 4
3. Solve:
   r = 2
4. Substitute:
   Coefficient = ⁶C₂ × 2²
5. Calculate:
   15 × 4 = 60

Final Result: Coefficient of x⁴ = 60

Middle Term In Binomial Expansion Class 11 Questions

Middle term depends on the number of terms. Since (a + b)ⁿ has n + 1 terms, odd and even values of n behave differently.

If n is even, there is one middle term. If n is odd, there are two middle terms.

Q26. What Is The Middle Term In Binomial Expansion?

The middle term is the central term in the expansion of (a + b)ⁿ.

If n is even, the middle term is Tₙ/₂₊₁. If n is odd, the middle terms are T₍ₙ₊₁₎/₂ and T₍ₙ₊₃₎/₂.

Example: (a + b)⁶ has 7 terms, so the middle term is T₄.

Q27. Find The Middle Term In (x + 2)⁶.

The middle term is 160x³.

1. Given expression:
   (x + 2)⁶
2. Number of terms:
   6 + 1 = 7
3. Middle term:
   T₄
4. Use formula:
   T₄ = ⁶C₃x³(2)³
5. Calculate:
   T₄ = 20 × 8 × x³ = 160x³

Final Result: 160x³

Q28. Find The Two Middle Terms In (a + b)⁵.

The two middle terms are 10a³b² and 10a²b³.

1. Given expression:
   (a + b)⁵
2. Number of terms:
   5 + 1 = 6
3. Middle terms:
   T₃ and T₄
4. Calculate:
   T₃ = ⁵C₂a³b² = 10a³b²
   T₄ = ⁵C₃a²b³ = 10a²b³

Final Result: 10a³b² and 10a²b³

Binomial Theorem Examples Class 11 For Numerical Evaluation

Numerical evaluation questions use nearby base numbers. NCERT uses examples like (98)⁵ to show why the theorem saves time.

The main trick is to write the number as 100 − 2, 100 + 2, or 1 − 0.01.

Q29. Evaluate (98)⁵ Using Binomial Theorem.

The value of (98)⁵ is 9039207968.

1. Write the number:
   98 = 100 − 2
2. Expand:
   (98)⁵ = (100 − 2)⁵
3. Use formula:
   (a − b)⁵ = a⁵ − 5a⁴b + 10a³b² − 10a²b³ + 5ab⁴ − b⁵
4. Substitute:
   = 100⁵ − 5(100⁴)(2) + 10(100³)(2²) − 10(100²)(2³) + 5(100)(2⁴) − 2⁵
5. Calculate:
   = 10000000000 − 1000000000 + 40000000 − 800000 + 8000 − 32
6. Simplify:
   = 9039207968

Final Result: 9039207968

Q30. Evaluate (96)³ Using Binomial Theorem.

The value of (96)³ is 884736.

1. Write the number:
   96 = 100 − 4
2. Expand:
   (96)³ = (100 − 4)³
3. Use formula:
   (a − b)³ = a³ − 3a²b + 3ab² − b³
4. Substitute:
   = 100³ − 3(100²)(4) + 3(100)(4²) − 4³
5. Calculate:
   = 1000000 − 120000 + 4800 − 64
6. Simplify:
   = 884736

Final Result: 884736

Q31. Evaluate (101)⁴ Using Binomial Theorem.

The value of (101)⁴ is 104060401.

1. Write the number:
   101 = 100 + 1
2. Expand:
   (101)⁴ = (100 + 1)⁴
3. Use formula:
   a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴
4. Substitute:
   = 100⁴ + 4(100³)(1) + 6(100²)(1²) + 4(100)(1³) + 1
5. Calculate:
   = 100000000 + 4000000 + 60000 + 400 + 1

Final Result: 104060401

Q32. Which Is Larger, (1.01)¹⁰⁰⁰⁰⁰⁰ Or 10000?

The number (1.01)¹⁰⁰⁰⁰⁰⁰ is larger than 10000.

1. Write expression:
   (1.01)¹⁰⁰⁰⁰⁰⁰ = (1 + 0.01)¹⁰⁰⁰⁰⁰⁰
2. Use first two terms:
   (1 + x)ⁿ = 1 + nx + positive terms
3. Substitute:
   = 1 + 1000000(0.01) + positive terms
4. Simplify:
   = 1 + 10000 + positive terms
5. Therefore:
   (1.01)¹⁰⁰⁰⁰⁰⁰ > 10000

Final Result: (1.01)¹⁰⁰⁰⁰⁰⁰ is larger

Binomial Theorem Approximation Questions

Approximation questions use the first few terms when the second part is small. This method works well for values near 1.

NCERT includes approximation-style practice through expressions like (0.99)⁵.

Q33. Find An Approximation Of (0.99)⁵ Using First Three Terms.

The approximate value of (0.99)⁵ is 0.9505.

1. Write the number:
   0.99 = 1 − 0.01
2. Use first three terms:
   (1 − x)⁵ ≈ 1 − 5x + 10x²
3. Substitute x = 0.01:
   (0.99)⁵ ≈ 1 − 5(0.01) + 10(0.01)²
4. Calculate:
   = 1 − 0.05 + 0.001
5. Simplify:
   = 0.951

Final Result: 0.951 approximately

Q34. Approximate (1.02)⁵ Using First Three Terms.

The approximate value of (1.02)⁵ is 1.104.

1. Write expression:
   (1.02)⁵ = (1 + 0.02)⁵
2. Use first three terms:
   (1 + x)⁵ ≈ 1 + 5x + 10x²
3. Substitute x = 0.02:
   = 1 + 5(0.02) + 10(0.02)²
4. Calculate:
   = 1 + 0.10 + 0.004

Final Result: 1.104 approximately

Q35. Why Are Later Terms Ignored In Binomial Approximation?

Later terms are ignored when the small number has high powers that become very small.

For x = 0.01, values like x³, x⁴, and x⁵ are tiny compared with x. Approximation keeps only terms with visible effect.

Example: (0.01)³ = 0.000001.

Binomial Theorem Divisibility Questions

Divisibility questions use a binomial form that contains the divisor. The expansion then separates a multiple of the divisor and a remainder.

NCERT uses this method to prove statements such as 6ⁿ − 5n leaves remainder 1 when divided by 25.

Q36. Prove That 6ⁿ − 5n Leaves Remainder 1 When Divided By 25.

The expression 6ⁿ − 5n leaves remainder 1 when divided by 25.

1. Write:
   6ⁿ = (1 + 5)ⁿ
2. Expand:
   (1 + 5)ⁿ = 1 + ⁿC₁5 + ⁿC₂5² + ⁿC₃5³ + ... + 5ⁿ
3. Since ⁿC₁ = n:
   6ⁿ = 1 + 5n + 25K, where K is an integer.
4. Subtract 5n:
   6ⁿ − 5n = 1 + 25K
5. Therefore:
   6ⁿ − 5n leaves remainder 1 when divided by 25.

Final Result: Remainder = 1

Q37. Prove That 9ⁿ⁺¹ − 8n − 9 Is Divisible By 64.

The expression 9ⁿ⁺¹ − 8n − 9 is divisible by 64.

1. Write:
   9ⁿ⁺¹ = 9(9ⁿ) = 9(1 + 8)ⁿ
2. Expand:
   (1 + 8)ⁿ = 1 + 8n + ⁿC₂8² + ⁿC₃8³ + ...
3. Multiply by 9:
   9ⁿ⁺¹ = 9 + 72n + 9[ⁿC₂8² + ⁿC₃8³ + ...]
4. Subtract:
   9ⁿ⁺¹ − 8n − 9 = 64n + 9[ⁿC₂8² + ⁿC₃8³ + ...]
5. Every remaining term has factor 64.

Final Result: 64 divides 9ⁿ⁺¹ − 8n − 9

Q38. Prove That a − b Is A Factor Of aⁿ − bⁿ.

The expression a − b is a factor of aⁿ − bⁿ.

1. Write:
   a = (a − b) + b
2. Then:
   aⁿ = [(a − b) + b]ⁿ
3. Expand:
   aⁿ = bⁿ + terms containing factor (a − b)
4. Subtract bⁿ:
   aⁿ − bⁿ = terms containing factor (a − b)

Final Result: a − b is a factor of aⁿ − bⁿ

NCERT Class 11 Maths Chapter 7 Questions On Combined Expressions

Combined expressions often use cancellation between (a + b)ⁿ and (a − b)ⁿ. Odd-power and even-power terms behave differently.

This pattern helps in questions like (a + b)⁴ − (a − b)⁴.

Q39. Find (a + b)⁴ − (a − b)⁴.

The value is 8a³b + 8ab³.

1. Expand first expression:
   (a + b)⁴ = a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴
2. Expand second expression:
   (a − b)⁴ = a⁴ − 4a³b + 6a²b² − 4ab³ + b⁴
3. Subtract:
   (a + b)⁴ − (a − b)⁴
4. Cancel like terms:
   = 8a³b + 8ab³

Final Result: 8a³b + 8ab³

Q40. Evaluate (√3 + √2)⁶ + (√3 − √2)⁶.

The value is 2702.

1. Let:
   a = √3, b = √2
2. Since addition removes odd-power terms:
   (a + b)⁶ + (a − b)⁶ = 2[a⁶ + 15a⁴b² + 15a²b⁴ + b⁶]
3. Substitute values:
   a² = 3, b² = 2
4. Calculate:
   a⁶ = 27
   15a⁴b² = 15 × 9 × 2 = 270
   15a²b⁴ = 15 × 3 × 4 = 180
   b⁶ = 8
5. Add inside bracket:
   27 + 270 + 180 + 8 = 485
6. Multiply by 2:
   2 × 485 = 970

Final Result: 970

Q41. Find (x + 1)⁶ + (x − 1)⁶.

The value is 2x⁶ + 30x⁴ + 30x² + 2.

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. Add both:
   Odd-power terms cancel.
4. Result:
   2x⁶ + 30x⁴ + 30x² + 2

Final Result: 2x⁶ + 30x⁴ + 30x² + 2

Q42. Evaluate (√2 + 1)⁶ + (√2 − 1)⁶.

The value is 198.

1. Use result:
   (x + 1)⁶ + (x − 1)⁶ = 2x⁶ + 30x⁴ + 30x² + 2
2. Put:
   x = √2
3. Values:
   x² = 2, x⁴ = 4, x⁶ = 8
4. Substitute:
   2(8) + 30(4) + 30(2) + 2
5. Calculate:
   16 + 120 + 60 + 2 = 198

Final Result: 198

Binomial Theorem Questions With Answers For Board Pattern

Board pattern questions often ask for a coefficient, one term, or a short proof. They test the formula more than long expansion.

Students should write a, b, n, and r before using the term formula.

Q43. Find The Coefficient Of x² In (1 + 3x)⁵.

The coefficient of x² is 90.

1. General term:
   Tᵣ₊₁ = ⁵Cᵣ(1)⁵⁻ʳ(3x)ʳ
2. For x²:
   r = 2
3. Coefficient:
   ⁵C₂ × 3²
4. Calculate:
   10 × 9 = 90

Final Result: 90

Q44. Find The Coefficient Of x³ In (2 + x)⁷.

The coefficient of x³ is 280.

1. General term:
   Tᵣ₊₁ = ⁷Cᵣ(2)⁷⁻ʳxʳ
2. For x³:
   r = 3
3. Coefficient:
   ⁷C₃ × 2⁴
4. Calculate:
   35 × 16 = 560

Final Result: 560

Q45. Find The Constant Term In (x + 1/x)⁶.

The constant term is 20.

1. General term:
   Tᵣ₊₁ = ⁶Cᵣx⁶⁻ʳ(1/x)ʳ
2. Simplify power of x:
   x⁶⁻ʳ × x⁻ʳ = x⁶⁻²ʳ
3. For constant term:
   6 − 2r = 0
4. Solve:
   r = 3
5. Constant term:
   ⁶C₃ = 20

Final Result: 20

Important Questions Class 11 Maths Chapter-Wise