If S is the sum of the first 10 terms of the series
tan ⁻¹( 1/3 ) + tan ⁻¹( 1/7 ) + tan ⁻¹( 1/13 ) + tan ⁻¹( 1/21 ) + ....
then tan(S) is equal to :

Solution

Maths sir · 4 steps

Write each term as a difference of two inverse tangents, and the whole sum telescopes away.

1. Spot the pattern. The denominators 3, 7, 13, 21, … are 1 + n(n+1) for n = 1, 2, 3, 4, …. So the nth term is tan⁻¹ 1/(1 + n(n+1)).
2. Split each term. Write the numerator 1 as (n+1) - n. This matches tan⁻¹ a - tan⁻¹ b = tan⁻¹(a-b)/(1+ab) with a = n+1 and b = n. The formula holds here because a and b are both positive.

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

3. Telescope. Add the terms from n = 1 to 10. Every middle term cancels, which leaves S = tan⁻¹ 11 - tan⁻¹ 1.
4. So. tan S = (11 - 1)/(1 + 11 × 1) = 10/12 = 5/6, which is option (d).

The idea to keep. When an inverse tangent has the form 1/(1+n(n+1)), write the 1 as (n+1)-n. The series then telescopes to the last term minus the first.

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