If Σr=1ⁿ Tᵣ=(2 n-1)(2 n+1)(2 n+3)(2 n+5)/64, then lim n → ∞ Σr=1ⁿ(1/Tᵣ) is equal to :

Solution

Maths sir · 4 steps

First pull out Tᵣ by subtracting consecutive sums, then split 1/Tᵣ so the series telescopes.

1. Find the general term. Use Tᵣ = Sᵣ - Sᵣ₋₁. The factors (2r-1)(2r+1)(2r+3) appear in both Sᵣ and Sᵣ₋₁. What is left is (2r+5)-(2r-3) = 8, so Tᵣ = 8(2r-1)(2r+1)(2r+3)/64. The options show the question wants this as the term for every r. Strictly, S₀ = -15/64, so T₁ itself would be 105/64, and that reading matches no option.

   Tᵣ = (2r-1)(2r+1)(2r+3)/8

2. Split for telescoping. The difference 1/(2r-1)(2r+1) - 1/(2r+1)(2r+3) equals 4/(2r-1)(2r+1)(2r+3). Since 1/Tᵣ is 8/(2r-1)(2r+1)(2r+3), that is 2 times the difference: 1/Tᵣ = 2[1/(2r-1)(2r+1) - 1/(2r+1)(2r+3)].
3. Add the terms. In the sum from r=1 to n, each negative part cancels the positive part of the next term. Only the first and last pieces survive: Σᵣ₌₁ⁿ1/Tᵣ = 2[1/(1 × 3) - 1/(2n+1)(2n+3)].
4. So. As n → ∞ the last piece goes to 0. The limit is 2 × 1/3 = 2/3, which is option (a).

The idea to keep. Given Sₙ, always find Tᵣ = Sᵣ - Sᵣ₋₁ first. For a product of terms in AP, write the reciprocal as the difference of two shorter products. Divide by the common difference of the outer factors.

Answer from the NTA final answer key. Solution by PrepWiser.