Solution:

To solve the recurrence equation T(2k) = 3 T(2k-1) - 1, we need to find a pattern in the terms and then find a closed-form solution.

Pattern of Terms:
Let's find the values of T(1), T(2), T(4), T(8), T(16), and T(32) to observe a pattern:

T(1) = 1
T(2) = 3 T(1) - 1 = 3(1) - 1 = 2
T(4) = 3 T(2) - 1 = 3(2) - 1 = 5
T(8) = 3 T(4) - 1 = 3(5) - 1 = 14
T(16) = 3 T(8) - 1 = 3(14) - 1 = 41
T(32) = 3 T(16) - 1 = 3(41) - 1 = 122

We can observe that the terms are increasing rapidly, but there doesn't seem to be a clear pattern.

Finding a Closed-Form Solution:
Let's try to find a closed-form solution for T(k) by substituting k = 2^m:

T(2^m) = 3 T(2^m-1) - 1

Let's divide both sides by T(2^m-1):

T(2^m) / T(2^m-1) = 3 - 1 / T(2^m-1)

Since T(2^m) / T(2^m-1) is a ratio of consecutive terms, it should approach a constant value as m increases.

Let's assume the ratio approaches a constant value R:

R = 3 - 1 / R

Simplifying the equation:

R^2 = 3 - 1

R^2 = 2

Taking the square root of both sides:

R = √2

Now, let's express T(2^m) in terms of m using the formula T(k) = R^(m-1) T(2):

T(2^m) = √2^(m-1) T(2)

Substituting k = 2^m:

T(k) = √2^(log2(k)-1) T(2)

Simplifying:

T(k) = 2^(log2(k)/2 - 1/2) T(2)

Since T(2) = 2 - 1 / 2 = 1/2:

T(k) = 2^(log2(k)/2 - 1/2) / 2 = (2^(log2(k)/2 - 1/2) - 1) / 2

Simplifying further:

T(k) = (2^(log2(k)/2) - 2^(-1/2)) / 2

T(k) = (2^(log2(k)/2) - 1) / 2

Finally, let's substitute k = 2k to get the solution in terms of k:

T(2k) = (2^(log2(2k)/2) - 1) / 2

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