Solve the recurrence T(n)=2T(n−1)+3nT(n) = 2T(n-1) + 3^nT(n)=2T(n−1)+3n with T(0)=1T(0) = 1T(0)=1. What is Θ(T(n))\Theta(T(n))Θ(T(n))?
Θ(3n)\Theta(3^n)Θ(3n)
Θ(2n)\Theta(2^n)Θ(2n)
Θ(n⋅2n)\Theta(n \cdot 2^n)Θ(n⋅2n)
Θ(n2)\Theta(n^2)Θ(n2)