Consider this recurrence relation: T(1) = 1 T(n) = 2 T(n/3) + 2n + 1 for n>1 The Master Theorem says that T（n）∈ Θ（）？

The Master Theorem can be applied to solve recurrence relations of the form T(n) = aT(n/b) + f(n), where a >= 1, b > 1, and f(n) is an asymptotically positive function.

In this case, a = 2, b = 3, and f(n) = 2n + 1.

The Master Theorem has three cases:

1. If f(n) = O(n^c), where c < log_b(a), then T(n) = Θ(n^log_b(a)).
2. If f(n) = Θ(n^c * log^k(n)), where c = log_b(a) and k >= 0, then T(n) = Θ(n^c * log^(k+1)(n)).
3. If f(n) = Ω(n^c), where c > log_b(a), if af(n/b) <= kf(n) for some constant k < 1 and sufficiently large n, then T(n) = Θ(f(n)).

In this case, f(n) = 2n + 1, which is Θ(n). The value of c is 1, which is less than log_b(a) = log_3(2). Therefore, we are in case 1 of the Master Theorem.

So, T(n) = Θ(n^log_b(a)) = Θ(n^log_3(2)).