If cot⁻¹(α) = cot⁻¹ 2 + cot⁻¹ 8 + cot⁻¹ 18 + cot⁻¹ 32 + ...... upto 100 terms, then α is :

Solution

Maths sir · 4 steps

Write each term as a difference of two tan inverses, and the 100-term sum collapses to two terms.

1. General term. The numbers are 2, 8, 18, 32, …, that is 2 × 1², 2 × 2², 2 × 3², …. So the nth term is cot⁻¹(2n²) = tan⁻¹ 1/2n² = tan⁻¹ 2/4n².
2. Split it. Write 2 = (2n+1) - (2n-1) and 4n² = 1 + (2n+1)(2n-1). Then the formula for tan⁻¹ x - tan⁻¹ y turns each term into a difference.

   tan⁻¹((2n+1)-(2n-1))/(1+(2n+1)(2n-1)) = tan⁻¹(2n+1) - tan⁻¹(2n-1)

3. Telescope. Adding from n = 1 to 100, every middle term cancels. Only tan⁻¹ 201 - tan⁻¹ 1 is left, which equals tan⁻¹(201-1)/(1+201) = tan⁻¹ 200/202 = tan⁻¹ 100/101.
4. So. cot⁻¹ α = tan⁻¹ 100/101 = cot⁻¹ 101/100, so α = 1.01. The answer is (c).

The idea to keep. When a sum of inverse tangents looks hopeless, write each term as tan⁻¹(x-y)/(1+xy) with consecutive x, y. It splits into tan⁻¹ x - tan⁻¹ y and the sum telescopes.

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