Exams › JEE Advanced › Maths

Evaluate the sum T = sum(r=0 to 20) r*(20-r)*[C(20,r)]².

1. 400 * C(38,18)
2. 400 * C(40,19)
3. 400 * C(39,19)
4. 400 * C(38,19)

Correct answer: 400 * C(38,18)

Solution

Each term equals 400*C(19,r-1)*C(19,r). Summing over r gives 400*sum C(19,r-1)*C(19,18-r+1)... applying Vandermonde: sum(r) C(19,r-1)*C(19,r) = C(38,18), so T = 400*C(38,18).

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 ]?
• What is the coefficient of x² in the expansion of (1 + x)⁵ * (1 + 2x)⁴?

⚔️ Practice JEE Advanced Maths free + battle 1v1 →