The sum of the infinite series cot ⁻¹(7/4)+cot ⁻¹(19/4)+cot ⁻¹(39/4)+cot ⁻¹(67/4)+…. is :

Solution

Maths sir · 4 steps

Find the general term, then split each inverse tan into a difference so the whole series telescopes.

1. General term. The numerators 7, 19, 39, 67 have differences 12, 20, 28, and the second differences are a constant 8. So the numerator is 4n² + 3, and the nth term is cot⁻¹((4n²+3)/4) = tan⁻¹(4/(4n²+3)).
2. Shape it for splitting. Divide top and bottom by 4 to get 1/(n² + 3/4) = 1/(1 + (n² - 1/4)) = ((n+1/2) - (n-1/2))/(1 + (n+1/2)(n-1/2)). This is the (x-y)/(1+xy) form.

   cot⁻¹((4n²+3)/4) = tan⁻¹(n+1/2) - tan⁻¹(n-1/2)

3. Telescope. Adding the first N terms, every middle term cancels and S_N = tan⁻¹(N+1/2) - tan⁻¹(1/2). Check with n = 1: tan⁻¹ 3/2 - tan⁻¹ 1/2 = tan⁻¹ 1/(1 + 3/4) = tan⁻¹ 4/7, which matches the first term.
4. So. As N → ∞, tan⁻¹(N+1/2) → π/2. The sum is π/2 - tan⁻¹(1/2), which is option (c).

The idea to keep. For a series of inverse cot or tan terms, write each term as tan⁻¹(x-y)/(1+xy) with x - y = 1. It splits into tan⁻¹ x - tan⁻¹ y and the series telescopes.

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