Let Σn=0^∞ (n³((2 n) !)+(2 n-1)(n !))/(n !)((2 n) !)=ae+b/e+c, where a, b, c ∈ ℤ and e=Σn=0^∞ 1/(n !) Then a²-b+c is equal to ____________.

Solution

Maths sir · 4 steps

Split the fraction into two sums: falling factorials handle one, an alternating series handles the other.

1. Split the term. Divide each part of the numerator by (n!)((2n)!). The general term becomes n³/n! + (2n-1)/(2n)!, so the series is two separate sums.
2. First sum. Write n³ = n(n-1)(n-2) + 3n(n-1) + n. Each piece cancels with n! and leaves a full series for e: Σ n(n-1)(n-2)/n! = e, Σ n(n-1)/n! = e, Σ n/n! = e. So Σ n³/n! = e + 3e + e = 5e.
3. Second sum. For n ≥ 1, 2n/(2n)! = 1/(2n-1)!, and at n = 0 that part is 0. So the second sum is the odd-factorial reciprocals minus the even-factorial reciprocals, which is an alternating series.

   Σₙ₌₀^∞(2n-1)/(2n)! = (1/1!+1/3!+…) - (1/0!+1/2!+…) = -Σₖ₌₀^∞(-1)ᵏ/k! = -1/e

4. So. The total is 5e - 1/e + 0, so a = 5, b = -1, c = 0. Then a² - b + c = 25 + 1 + 0 = 26.

The common slip: 24. Taking b = +1 instead of -1 (losing the minus sign on 1/e) gives 25 - 1 + 0 = 24, not 26.

The idea to keep. For Σ p(n)/n!, rewrite p(n) in falling factorials so each piece becomes a multiple of e. Odd-factorial reciprocals minus even-factorial reciprocals always give -1/e.

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