Let Sₙ = 1 . (n - 1) + 2 . (n - 2) + 3 . (n - 3) + ..... + (n - 1) . 1, n ≥ 4.
The sum Σₙ₌₄^∞ ( 2Sₙ/n! - 1/(n - 2)! ) is equal to :

Solution

Maths sir · 4 steps

Find Sₙ in closed form, and each term collapses to 1/3(n-3)!, a piece of e.

1. Closed form for Sn. The general term is r(n-r) for r = 1 to n-1. Split it as nr - r² and use the standard sums of r and r².

   Sₙ = n·(n-1)n/2 - (n-1)n(2n-1)/6 = n(n-1)(3n-(2n-1))/6 = (n-1)n(n+1)/6

2. Simplify each term. Since n! = n(n-1)(n-2)!, we get 2Sₙ/n! = (n+1)/3(n-2)!. Subtracting 1/(n-2)! = 3/3(n-2)! leaves (n-2)/3(n-2)! = 1/3(n-3)!.
3. Quick check. At n = 4, S₄ = 3 + 4 + 3 = 10, and the term is 20/24 - 1/2 = 1/3. This matches 1/(3 · 1!).
4. So. Put k = n-3, so the sum is 1/3(1/1! + 1/2! + 1/3! + …) = 1/3(e-1). The answer is (a), (e-1)/3.

The idea to keep. When a series has factorials in the denominator, cancel until each term looks like 1/k!. Then match it to e = Σₖ₌₀^∞ 1/k! and watch where k starts.

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