RowQ
The Vault
RowQ
The Vault
CBSE Class 11 Maths · 10 questions · 24 marks
Multiplying out (x + y) even ten times by hand is unthinkable, yet the binomial theorem hands you every term of the expansion in one clean formula built from combinations. This chapter teaches you to write the general term, spot the middle term, and pull out a specific coefficient without ever multiplying brackets together. It is also your first serious workout with nCr notation, which reappears constantly in probability later.
The number of terms in the expansion of (2x - 3)⁹ is:
Answer
The number of terms is 10. For (x + y)ⁿ, the expansion always has n + 1 terms, since r runs from 0 to n. Here n = 9, so the count is 9 + 1 = 10 terms.
The coefficient of x⁵ in the expansion of (1 + x)⁸ is:
Answer
The coefficient is 56. The general term of (1 + x)⁸ is Tᵣ₊₁ = 8Cr xʳ, so x⁵ appears when r = 5. The coefficient is 8C5 = 8! / (5!3!) = (8 × 7 × 6) / (3 × 2 × 1) = 336/6 = 56.
The middle term(s) of the expansion of (x + y)⁶ occur at position:
Answer
The middle term is the 4th term only. Since n = 6 is even, there is exactly one middle term, given by the ((n/2)+1)-th term. That is the ((6/2)+1)-th = 4th term.
The sum nC0 + nC1 + nC2 + ... + nCn equals:
Answer
The sum equals 2ⁿ. Setting x = 1 in (1 + x)ⁿ = Σ nCr xʳ gives (1+1)ⁿ = Σ nCr, so 2ⁿ = nC0 + nC1 + ... + nCn.
Assertion (A): The coefficient of x³ in the expansion of (1 + x)⁷ equals the coefficient of x⁴ in the same expansion. Reason (R): nCr = nC(n-r) for every valid r.
Answer
Both A and R are true and R is the correct explanation of A. The coefficient of x³ is 7C3 and the coefficient of x⁴ is 7C4. Since 7 - 3 = 4, the symmetry property 7C3 = 7C4 applies directly, so 7C3 = 35 = 7C4, and Assertion A is true. Reason R is the standard symmetry identity nCr = nC(n-r), and it is exactly this identity that forces the two coefficients to be equal, so R correctly explains A.
Write the general term in the expansion of (x - 2y)¹⁰ and use it to find the 5th term.
Answer
The general term is Tᵣ₊₁ = 10Cr x¹⁰⁻ʳ (-2y)ʳ. For the 5th term, r + 1 = 5, so r = 4. T₅ = 10C4 x⁶ (-2y)⁴ = 210 · x⁶ · 16y⁴. So T₅ = 3360 x⁶y⁴.
Find the coefficient of x⁷ in the expansion of (x² + 3/x)⁸.
Answer
The general term is Tᵣ₊₁ = 8Cr (x²)⁸⁻ʳ (3/x)ʳ = 8Cr 3ʳ x¹⁶⁻²ʳ⁻ʳ = 8Cr 3ʳ x¹⁶⁻³ʳ. For the power of x to equal 7, set 16 - 3r = 7, giving 3r = 9, so r = 3. The coefficient is 8C3 · 3³ = 56 × 27 = 1512.
Using the binomial theorem, expand (2x + 1)⁵ completely, and hence evaluate (2.01)⁵ correct to three decimal places by writing 2.01 = 2 + 0.01.
Answer
Step 1 — Expand (2x + 1)⁵ using Tᵣ₊₁ = 5Cr (2x)⁵⁻ʳ (1)ʳ: r=0: 5C0(2x)⁵ = 32x⁵ r=1: 5C1(2x)⁴ = 5×16x⁴ = 80x⁴ r=2: 5C2(2x)³ = 10×8x³ = 80x³ r=3: 5C3(2x)² = 10×4x² = 40x² r=4: 5C4(2x) = 5×2x = 10x r=5: 5C5 = 1 So (2x+1)⁵ = 32x⁵ + 80x⁴ + 80x³ + 40x² + 10x + 1. Step 2 — To evaluate (2.01)⁵, write 2.01 = 2 + 0.01 and expand (2 + 0.01)⁵ directly using Tᵣ₊₁ = 5Cr 2⁵⁻ʳ (0.01)ʳ: r=0: 32 r=1: 5×16×0.01 = 0.8 r=2: 10×8×0.0001 = 0.008 r=3 onward: powers of (0.01)³ and higher are under 0.000001, negligible to 3 decimals. Step 3 — Sum: 32 + 0.8 + 0.008 = 32.808. So (2.01)⁵ ≈ 32.808.
If the coefficients of the (2r+4)-th and (r-2)-th terms in the expansion of (1 + x)¹⁸ are equal, find r.
Answer
Step 1 — The coefficient of the (k+1)-th term of (1+x)¹⁸ is 18Ck. For the (2r+4)-th term, k = 2r+3, so its coefficient is 18C(2r+3). For the (r-2)-th term, k = r-3, so its coefficient is 18C(r-3). Step 2 — Set them equal: 18C(2r+3) = 18C(r-3). Step 3 — Two binomial coefficients nCa = nCb are equal when a = b or a + b = n. Case 1: 2r+3 = r-3 gives r = -6, rejected since r must be a positive integer with valid term positions. Case 2: (2r+3) + (r-3) = 18 gives 3r = 18, so r = 6. Step 4 — Check: 2r+3 = 15 and r-3 = 3, and indeed 18C15 = 18C3 by symmetry (18-15=3). Valid. So r = 6.
Read the following and answer the questions that follow: A student is studying the expansion of (a + b)ⁿ for a project. She fixes a = x, b = 1 and n = 7, and wants to explore individual terms and coefficients of (x + 1)⁷ without fully expanding it. (a) Write the general term Tᵣ₊₁ of the expansion. (b) Find the 3rd term of the expansion. (c) Find the coefficient of x⁴ in the expansion. (d) Find the sum of all the binomial coefficients in this expansion.
Answer
(a) Tᵣ₊₁ = 7Cr x⁷⁻ʳ (1)ʳ = 7Cr x⁷⁻ʳ. (b) For the 3rd term, r+1 = 3, so r = 2: T₃ = 7C2 x⁵ = 21x⁵. (c) The coefficient of x⁴ occurs when 7-r = 4, so r = 3: coefficient = 7C3 = 35. (d) Sum of binomial coefficients = 7C0+7C1+...+7C7 = 2⁷ = 128.
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