Exams › JEE Advanced › Maths

If (1 + x + x²)¹⁰ = sum(r=0 to 20) a_r * x^r, then the value of a₅ - (10C1)*a₄ + (10C2)*a₃ - (10C3)*a₂ + (10C4)*a₁ - (10C5)*a₀ is equal to:

1. 0
2. -1
3. 1
4. 10C5

Correct answer: 0

Solution

Let f(x) = (1+x+x²)¹⁰ = sum a_r * x^r. Consider g(x) = (1-x)¹⁰ * (1+x+x²)¹⁰ = ((1-x)(1+x+x²))¹⁰ = (1 - x³)¹⁰. The coefficient of x⁵ in (1-x³)¹⁰ = sum C(10,k)*(-1)^k * x^(3k). For x⁵: need 3k=5, which gives k=5/3 (not integer). So coefficient of x⁵ in (1-x³)¹⁰ is 0. But the coefficient of x⁵ in (1-x)¹⁰ * f(x) = sumₖ₌₀⁵ (-1)^k * C(10,k) * a₅₋ₖ. Since this equals the coefficient of x⁵ in (1-x³)¹⁰ = 0. Answer: 0.

Related JEE Advanced Maths questions

• In the expansion of (1 + x)ⁿ, let the binomial coefficients be represented as C₀, C₁, C₂,..., Cₙ. If p and q are such that p + q = 1, what is the value of Σᵣ₌₀ nCr pʳqⁿ⁻ʳ?
• What is the result of the summation Σ₀≤k≤n ∑ᵢ i * Cᵢ?
• The expression E5(n) is defined as Σ (nCk × k⁴) / Σ (nCk × k³), and it simplifies to n / 4. For n ≤ 109, E5(n) is less than 26. Therefore, the smallest value of n that satisfies the condition E5(n) ≥ 26 is one of the following:
• What is the coefficient of x¹¹ in the expansion of (1 + x²)⁴ (1 + x³)⁷ (1 + x)¹²?
• What is the coefficient of x⁵⁰ in the expansion of the sum S = sum(r=0 to 100) [ C(100,r) * (x-3)^(100-r) * 2^r ]?
• Evaluate the sum T = sum(r=0 to 20) r*(20-r)*[C(20,r)]².

⚔️ Practice JEE Advanced Maths free + battle 1v1 →