JEE Main 2023 · 1 February · morning · Single correct · +4 / −1

The sum of 10 terms of the series
1/(1 + 1² + 1⁴) + 2/(1 + 2² + 2⁴) + 3/(1 + 3² + 3⁴) + .... is

  1. (a) 58/111
  2. (b) 56/111
  3. (c) 55/111
  4. (d) 59/111
Answer and solution

Answer (c) 55/111

Solution

Maths sir · 4 steps

Factor the denominator as a difference of squares and each term splits into two pieces that cancel.

  1. Factor the denominator. Write 1 + n² + n⁴ = (n²+1)² - n², a difference of squares. So it equals (n² - n + 1)(n² + n + 1).
  2. Split the term. The two factors differ by (n²+n+1) - (n²-n+1) = 2n, and the numerator is n, half of that gap. So each term breaks into a difference of two fractions.

    n/(1+n²+n⁴) = 1/2(1/(n²-n+1) - 1/(n²+n+1))

  3. Watch it telescope. Since n² + n + 1 = (n+1)² - (n+1) + 1, the second fraction of term n is the first fraction of term n+1. The denominators run 1, 3, 7, 13, …, 111, and every middle fraction cancels.
  4. So. Only the first and last fractions survive: S₁₀ = 1/2(1/1 - 1/(10²+10+1)) = 1/2 × 110/111 = 55/111, which is option (c).

The idea to keep. When you see 1 + n² + n⁴, think (n²+1)² - n². If the two factors differ by a multiple of the numerator, split into partial fractions and let the series telescope.

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

Study Sequences & Series

Asked about 250 times in JEE Main since 2020. This question is from 2023.