CBSE Important Questions Class 11 Maths Chapter 7 – Binomial Theorem

Class 11 Maths Chapter 7 Binomial Theorem is part of the CBSE Class 11 Maths syllabus. It gives a quick way to expand (a + b)ⁿ without multiplying the brackets again and again. Students learn Pascal's triangle, the general term of the expansion, how to find a particular term such as the term containing x⁵, the middle term, and the term independent of x.

These CBSE important questions for Class 11 Maths Chapter 7 cover writing the binomial expansion, finding the general term, locating the middle term, finding the coefficient of a given power of x, the term independent of x, and using the expansion to find the value of numbers such as (98)³ and (1.01)¹⁰. Every question has a clear step-by-step solution in simple English, so students can practise on their own, parents can check the working, and teachers can use them for class tests.

Q1. State the binomial theorem for a positive integral index. Write the general term of the expansion of (a + b)ⁿ.

Answer: Binomial theorem: For any positive integer n,
(a + b)ⁿ = nC₀ aⁿ + nC₁ aⁿ⁻¹b + nC₂ aⁿ⁻²b² + ... + nCₙ bⁿ
General term: The (r + 1)th term is
T_r+1 = nC_r a^n−r b^r
There are always (n + 1) terms in the expansion, and the sum of the powers of a and b in every term is n.

Q2. Expand (x + 2y)⁵ using the binomial theorem.

Answer: Using (a + b)⁵ with a = x and b = 2y, and the coefficients 1, 5, 10, 10, 5, 1:
= x⁵ + 5x⁴(2y) + 10x³(2y)² + 10x²(2y)³ + 5x(2y)⁴ + (2y)⁵
= x⁵ + 10x⁴y + 10x³(4y²) + 10x²(8y³) + 5x(16y⁴) + 32y⁵
= x⁵ + 10x⁴y + 40x³y² + 80x²y³ + 80xy⁴ + 32y⁵

Q3. Find the 4th term in the expansion of (2x − 3y)⁷.

Answer: Using T_r+1 = nC_r a^n−r b^r with n = 7, a = 2x, b = −3y.
For the 4th term, r + 1 = 4, so r = 3.
T₄ = ⁷C₃ (2x)⁴ (−3y)³
= 35 × 16x⁴ × (−27y³)
= 35 × 16 × (−27) x⁴y³
= −15120 x⁴y³

Q4. Find the middle term in the expansion of (x + 1/x)¹⁰.

Answer: Here n = 10, which is even, so there is only one middle term: the (n/2 + 1)th term, that is the 6th term.
T₆ = ¹⁰C₅ (x)⁵ (1/x)⁵
= 252 × x⁵ × x⁻⁵
= 252 × x⁰
= 252

Q5. Find the term independent of x in the expansion of (x + 1/x²)⁹.

Answer: The general term is
T_r+1 = ⁹C_r (x)^9−r (1/x²)^r
= ⁹C_r x^9−r x^−2r
= ⁹C_r x^9−3r
For the term to be independent of x, the power must be zero:
9 − 3r = 0, so r = 3
T₄ = ⁹C₃ = (9 × 8 × 7)/(3 × 2 × 1) = 84

Q6. Find the coefficient of x⁵ in the expansion of (x + 3)⁸.

Answer: The general term is
T_r+1 = ⁸C_r x^8−r 3^r
For the power of x to be 5:
8 − r = 5, so r = 3
T₄ = ⁸C₃ x⁵ 3³
= 56 × 27 × x⁵
So the coefficient of x⁵ is 1512.

Q7. Using the binomial theorem, evaluate (98)³.

Answer: Write 98 as (100 − 2), which makes the calculation easy.
(100 − 2)³ = 100³ − 3(100)²(2) + 3(100)(2)² − 2³
= 1000000 − 3(10000)(2) + 3(100)(4) − 8
= 1000000 − 60000 + 1200 − 8
= 941192

Q8. Which is larger, (1.01)¹⁰⁰⁰⁰⁰ or 10,000?

Answer: Write 1.01 as (1 + 0.01) and use the binomial expansion:
(1 + 0.01)¹⁰⁰⁰⁰⁰ = 1 + 100000(0.01) + (terms that are all positive)
= 1 + 1000 + positive terms
Just the third term alone is
¹⁰⁰⁰⁰⁰C₂(0.01)² = [100000 × 99999 / 2] × 0.0001, which is already about 499995.
So the total is far more than 10,000.
Therefore (1.01)¹⁰⁰⁰⁰⁰ is larger.

Q9. Find the number of terms in the expansion of (2x − 3y)¹⁵ and write its last term.

Answer: In the expansion of (a + b)ⁿ there are always (n + 1) terms.
Here n = 15, so the number of terms is 15 + 1 = 16.
The last term is obtained when r = n = 15:
T₁₆ = ¹⁵C₁⁵ (2x)⁰ (−3y)¹⁵
= 1 × 1 × (−3)¹⁵ y¹⁵
= −3¹⁵ y¹⁵

Q10. Write the first four rows of Pascal's triangle and explain how it relates to the binomial coefficients.

Answer: Pascal's triangle is built by writing 1 at both ends of each row, and every other number is the sum of the two numbers just above it.
Row for n = 0:  1
Row for n = 1:  1  1
Row for n = 2:  1  2  1
Row for n = 3:  1  3  3  1
Row for n = 4:  1  4  6  4  1
Relation: The numbers in the row for n are exactly nC₀, nC₁, nC₂, ..., nCₙ, which are the coefficients in the expansion of (a + b)ⁿ. For example, from the n = 3 row we get (a + b)³ = a³ + 3a²b + 3ab² + b³.

Q11. Prove that the sum of the binomial coefficients in the expansion of (1 + x)ⁿ is 2ⁿ.

Answer: The expansion is
(1 + x)ⁿ = nC₀ + nC₁x + nC₂x² + ... + nCₙxⁿ
This is true for every value of x, so we may put x = 1:
(1 + 1)ⁿ = nC₀ + nC₁ + nC₂ + ... + nCₙ
So the sum of all the binomial coefficients is 2ⁿ. Hence proved.
(Putting x = −1 similarly shows that the alternating sum is 0.)

Q12. Find the middle terms in the expansion of (3x − x³/6)⁷.

Answer: Here n = 7, which is odd, so there are two middle terms: the 4th and the 5th.
4th term (r = 3):
T₄ = ⁷C₃ (3x)⁴ (−x³/6)³
= 35 × 81x⁴ × (−x⁹/216)
= −(35 × 81 / 216) x¹³ = −(105/8) x¹³
5th term (r = 4):
T₅ = ⁷C₄ (3x)³ (−x³/6)⁴
= 35 × 27x³ × (x¹²/1296)
= (35 × 27 / 1296) x¹⁵ = (35/48) x¹⁵

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