The sum of the series
1/(x + 1) + 2/(x² + 1) + 2²/(x⁴ + 1) + ...... + 2¹⁰⁰/(x^(2¹⁰⁰) + 1) when x = 2 is :

Solution

Maths sir · 4 steps

Put 1/(x-1) in front and watch each term swallow the next, like dominoes falling.

1. The collapsing identity. Write y = x^(2ᵏ). Then 2ᵏ/(y-1) - 2ᵏ/(y+1) = (2ᵏ × 2)/(y²-1) = 2ᵏ⁺¹/(x^(2ᵏ⁺¹)-1). Subtracting the k-th term of the series from 2ᵏ/(x^(2ᵏ)-1) gives the same kind of expression one step higher.
2. Run the chain. Start from 1/(x-1) and subtract the terms one by one. Subtracting 1/(x+1) leaves 2/(x²-1). Subtracting 2/(x²+1) leaves 4/(x⁴-1). After all 101 terms, from k = 0 to k = 100, only one piece is left.

   1/(x-1) - S = 2¹⁰¹/(x^(2¹⁰¹)-1) ⇒ S = 1/(x-1) - 2¹⁰¹/(x^(2¹⁰¹)-1)

3. Put x = 2. 1/(x-1) = 1, so S = 1 - 2¹⁰¹/(2^(2¹⁰¹)-1). The sum is just below 1, which fits: 1/3 + 2/5 + 4/17 + … climbs towards 1 but never reaches it.
4. So. The answer is (d): 1 minus 2¹⁰¹ over a power of 2 minus 1. The power should be 2^(2¹⁰¹); the printed 2⁴⁰⁰ is a misprint of that exponent.

The idea to keep. When the denominators go x+1, x²+1, x⁴+1, …, add 1/(x-1) in front. Then (a-1)(a+1) = a²-1 makes the whole series collapse to its first and last pieces.

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