Exams › JEE Advanced › Maths

Find the value of the sum: 1/(1! * 50!) + 1/(3! * 48!) + 1/(5! * 46!) +... + 1/(49! * 2!) + 1/(51! * 0!). Express your answer in the form 2ⁿ / m!.

1. 2⁵⁰ / 50!
2. 2⁵⁰ / 51!
3. 2⁵¹ / 51!
4. 2⁵¹ / 50!

Correct answer: 2⁵⁰ / 51!

Solution

The sum equals (1/51!) * [C(51,1) + C(51,3) +... + C(51,51)] = (1/51!) * 2⁵⁰ = 2⁵⁰/51!.

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 →