The value of 1 + 1/(1 + 2) + 1/(1 + 2 + 3) + .... + 1/(1 + 2 + 3 + ..... + 11) is equal to:

Solution

Maths sir · 4 steps

Each denominator is a triangular number, so every term splits into a difference of two fractions.

1. General term. The nth denominator is 1 + 2 + … + n = n(n+1)/2. So the nth term is 2/n(n+1), for n = 1 to 11.
2. Split it. Use partial fractions: 1/n(n+1) = 1/n - 1/(n+1). Now each term is 2(1/n - 1/(n+1)).

   Σₙ₌₁¹¹ 2(1/n - 1/(n+1)) = 2(1 - 1/2 + 1/2 - 1/3 + … + 1/11 - 1/12)

3. Telescope. Every middle fraction cancels with its neighbour. Only the first and last pieces survive, so the sum is 2(1 - 1/12) = 2 × 11/12.
4. So. The value is 22/12 = 11/6, which is option (b).

The idea to keep. When a term looks like 1/n(n+1), split it into 1/n - 1/(n+1). The middle terms cancel, and only the first piece and the very last piece are left.

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