Correct : a

Given: T(n) = 2T(n − 1) + n2ⁿ,   T(0) = 1

We need to find the asymptotic bound Θ.

Expand the recurrence
T(n) = 2T(n−1) + n2ⁿ
= 2(2T(n−2) + (n−1)2ⁿ⁻¹) + n2ⁿ
= 4T(n−2) + 2(n−1)2ⁿ⁻¹ + n2ⁿ
= 4T(n−2) + (n−1)2ⁿ + n2ⁿ
= 4T(n−2) + (2n − 1)2ⁿ

T(n) = 2^kT(n−k) + Σ_i=0^k−1 2ⁱ(n−i)2ⁿ⁻ⁱ

When k = n:
T(n) = 2ⁿT(0) + Σ_i=0ⁿ⁻¹ 2ⁱ(n−i)2ⁿ⁻ⁱ

Simplify the terms
Since 2ⁱ × 2ⁿ⁻ⁱ = 2ⁿ,
T(n) = 2ⁿ + 2ⁿ Σ_i=0ⁿ⁻¹ (n−i)

Σ_i=0ⁿ⁻¹ (n−i) = n + (n−1) + (n−2) + … + 1 = n(n + 1)/2
T(n) = 2ⁿ + 2ⁿ × (n(n + 1)/2)

T(n) = 2ⁿ (1 + n(n + 1)/2)

As n grows large, dominant term ≈ (n² × 2ⁿ).

Therefore: T(n) = Θ(n² × 2ⁿ)

Final Answer:
Option A: T(n) = Θ(n² × 2ⁿ)