JEE Main 2021 · 26 February · evening · Single correct · +4 / −1

The sum of the series
Σₙ₌₁^∞ (n² + 6n + 10)/(2n + 1)! is equal to :

  1. (a) 41/8e + 19/8e⁻¹ - 10
  2. (b) 41/8e - 19/8e⁻¹ - 10
  3. (c) 41/8e + 19/8e⁻¹ + 10
  4. (d) - 41/8e + 19/8e⁻¹ - 10
Answer and solution

Answer (b) 41/8e - 19/8e⁻¹ - 10

Solution

Maths sir · 3 steps

Rewrite the top in terms of m = 2n+1; then every piece becomes an odd or even half of e.

  1. Change the variable. Put m = 2n+1, so n = (m-1)/2 and m runs over 3, 5, 7, …. Then n² + 6n + 10 = (m-1)²/4 + 3(m-1) + 10 = (m² + 10m + 29)/4.
  2. Split the top. Write m² + 10m + 29 = m(m-1) + 11m + 29. Each part cancels against m!, so the general term becomes 1/4[1/(m-2)! + 11/(m-1)! + 29/m!].
    PieceTerms (m = 3, 5, 7, …)Value
    Σ 1/(m-2)!1/1! + 1/3! + 1/5! + …(e - e⁻¹)/2
    Σ 1/(m-1)!1/2! + 1/4! + 1/6! + …(e + e⁻¹)/2 - 1
    Σ 1/m!1/3! + 1/5! + 1/7! + …(e - e⁻¹)/2 - 1

    4S = 30·(e - e⁻¹)/2 + 11·(e + e⁻¹)/2 - 11 - 29 = 41/2e - 19/2e⁻¹ - 40

  3. So. Dividing by 4, S = 41/8e - 19/8e⁻¹ - 10 ≈ 3.06, which is option (b). As a check, the first two terms alone give 17/6 + 26/120 ≈ 3.05.

The idea to keep. With (2n+1)! or (2n)! below, rewrite the numerator in falling factorials of that same number. The sums then reduce to (e ± e⁻¹)/2, minus whichever starting terms are missing.

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

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Asked about 250 times in JEE Main since 2020. This question is from 2021.