Exams › JEE Advanced › Maths
In the expansion of (x - 1/x)⁶, the term independent of x is
- -20
- 20
- 30
- -30
Correct answer: -20
Solution
General term T_(r+1) = C(6,r)*(x)^(6-r)*(-1/x)^r = C(6,r)*(-1)^r * x^(6-2r). For independence from x: 6-2r = 0 => r = 3. Term = C(6,3)*(-1)³ = 20*(-1) = -20.
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 →