The sum Σₙ₌₁^∞ (2n² + 3n + 4)/(2n)! is equal to :

Solution

Maths sir · 4 steps

Rewrite the numerator using 2n(2n-1) and 2n, so each piece cancels against the (2n)! below it.

1. Reshape the numerator. Put m = 2n. Then 2n² = m²/2 = 1/2m(m-1) + m/2 and 3n = 3m/2. So 2n² + 3n + 4 = 1/2 (2n)(2n-1) + 2(2n) + 4.

   (2n²+3n+4)/(2n)! = 1/2·1/(2n-2)! + 2/(2n-1)! + 4/(2n)!

2. Use the e-series. Adding and subtracting e = Σ 1/k! and e⁻¹ = Σ (-1)ᵏ/k! gives the even and odd parts. For n ≥ 1: Σ 1/(2n-2)! = (e + e⁻¹)/2, Σ 1/(2n-1)! = (e - e⁻¹)/2, and Σ 1/(2n)! = (e + e⁻¹)/2 - 1, because the 1/0! term is missing.
3. Combine. The sum is (e + e⁻¹)/4 + (e - e⁻¹) + 2(e + e⁻¹) - 4. The coefficient of e is 1/4 + 1 + 2 = 13/4, and the coefficient of 1/e is 1/4 - 1 + 2 = 5/4.
4. So. The sum equals 13e/4 + 5/4e - 4, which is option (b).

The idea to keep. With a polynomial over a factorial, write the polynomial in falling products like m(m-1), m, 1 to match the factorial. Then watch where each series starts.

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