Class 11 Maths Chapter 7 Binomial Theorem – Extra Questions with Answers

Extra practice questions for Class 11 Maths Chapter 7 (Binomial Theorem), beyond the textbook. These Class 11 Maths Chapter 7 important questions are handy for last-minute exam practice.

Very Short Answer Questions (1 mark)

Q1. How many terms are in the expansion of (x+y)⁵?
Ans: 8 terms (7+1).

Q2. Find ⁴C₂, the coefficient of the middle term in (a+b)⁴.
Ans: 4!/[2!2!] = 6.

Q3. Expand (a+b)² using the binomial theorem.
Ans: a²+2ab+b².

Q4. What is the sum of binomial coefficients for (a+b)⁴?
Ans: 2⁴ = 16.

Q5. In (a+b)⁵, is ⁷C₂ equal to ⁷C₅?
Ans: Yes, by the symmetry property ⁿC_k=ⁿC_n−k (7−2=5).

Short Answer Questions (2–3 marks)

Q6. Expand (x+2)³ using the binomial theorem, showing each term.
Ans: ³C₀x³+³C₁x²(2)+³C₂x(2)²+³C₃(2)³ = x³+3(2)x²+3(4)x+8 = x³+6x²+12x+8.

Q7. Find the 4th term in the expansion of (x+y)⁵, using the general term formula.
Ans: T₄ = T₃₊₁, so k=3: T₄ = ⁷C₃ x⁷⁻³ y³ = 35x⁴y³.

Q8. Find the middle term(s) of the expansion of (a+b)⁶ (n=6, even).
Ans: Since n=6 is even, there is one middle term: the ((6/2)+1) = 4th term. T₄ = ⁶C₃ a³b³ = 20a³b³.

Higher-Order Thinking / Application Questions

Q9. Using the binomial theorem, find the value of (1.02)⁴ correct to 4 decimal places, by expressing 1.02 as (1+0.02) and using only the first three terms (ignoring higher-order terms as negligible).
Ans: (1+0.02)⁴ ≈ ⁴C₀(1)⁴ + ⁴C₁(1)³(0.02) + ⁴C₂(1)²(0.02)² = 1 + 4(0.02) + 6(0.0004) = 1 + 0.08 + 0.0024 = 1.0824.

Q10. Explain why the coefficients in row n of Pascal’s Triangle sum to 2ⁿ, connecting this to the binomial theorem with a=b=1.
Ans: Substituting a=1, b=1 into the binomial theorem: (1+1)ⁿ = ∑ ⁿC_k(1)^n−k(1)^k = ∑ ⁿC_k = ⁿC₀+ⁿC₁+…+ⁿC_n. Since (1+1)ⁿ = 2ⁿ, this directly shows that the sum of all binomial coefficients (the entries in row n of Pascal’s Triangle) must equal 2ⁿ.

Written by Satish

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