Let Sₖ = Σᵣ₌₁ᵏ tan ⁻¹( 6ʳ/(2²ʳ⁺¹ + 3²ʳ⁺¹) ) . Then lim _(k → ∞) Sₖ is equal to :

Solution

Maths sir · 4 steps

Rewrite each term as a difference of two inverse tangents and the whole sum telescopes.

1. Tidy the term. Divide the top and bottom by 2 · 4ʳ and put x = (3/2)ʳ. The top 6ʳ becomes x/2 and the bottom 2 · 4ʳ + 3 · 9ʳ becomes 1 + 3/2x².
2. Spot the pattern. The bottom is 1 + (3/2)ʳ⁺¹(3/2)ʳ. The top is (3/2)ʳ⁺¹ - (3/2)ʳ = x/2. So each term has the form tan⁻¹(a-b)/(1+ab).

   tan⁻¹(6ʳ/(2²ʳ⁺¹+3²ʳ⁺¹)) = tan⁻¹(3/2)ʳ⁺¹ - tan⁻¹(3/2)ʳ

3. Add and cancel. Each term cancels part of the next one. Only the last piece and the first piece survive: Sₖ = tan⁻¹(3/2)ᵏ⁺¹ - tan⁻¹ 3/2.
4. So. As k → ∞, (3/2)ᵏ⁺¹ → ∞, so the first part tends to π/2. The limit is π/2 - tan⁻¹ 3/2 = cot⁻¹ 3/2, which is option (a).

The idea to keep. When a sum of tan⁻¹ terms appears, try to write each term as tan⁻¹(a-b)/(1+ab), with a and b consecutive terms of a pattern. The sum then telescopes.

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