Let a=1+(² C₂)/(3 !)+(³ C₂)/(4 !)+(⁴ C₂)/(5 !)+...., b=1+(¹ C₀+ ¹ C₁)/(1 !)+(² C₀+ ² C₁+ ² C₂)/(2 !)+(³ C₀+ ³ C₁+ ³ C₂+ ³ C₃)/(3 !)+.... Then (2 b)/a² is equal to _________.

Solution

Maths sir · 4 steps

Turn each series into pieces of eˣ: a comes out as e/2 and b as e².

1. General term of a. After the leading 1, the term is ⁿC₂/(n+1)! = n(n-1)/(2 (n+1)!) for n ≥ 2. Split n(n-1) so that each piece cancels against the factorial: n(n-1) = (n+1)n - 2(n+1) + 2.

   n(n-1)/(n+1)! = 1/(n-1)! - 2/n! + 2/(n+1)!

2. Sum the tails. Summing from n = 2, the three tails are Σ 1/(n-1)! = e - 1, Σ 2/n! = 2(e-2) and Σ 2/(n+1)! = 2(e - 5/2) = 2e - 5. Together they give (e-1) - 2(e-2) + (2e-5) = e - 2, so a = 1 + (e-2)/2 = e/2.
3. Find b. Each numerator is a full row sum: ⁿC₀ + ⁿC₁ + … + ⁿCₙ = 2ⁿ. So b = Σ_(n ≥ 0) 2ⁿ/n! = e², with the leading 1 being the n = 0 term.
4. So. 2b/a² = 2e²/(e²/4) = 8.

The idea to keep. When a factorial sits in the denominator, rewrite the polynomial on top in falling pieces like (n+1)n and (n+1). Each piece cancels into a 1/k! tail of e.

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