Final answer:
By using iteration, you can find that the time complexity is O(7^log2(n)).
Step-by-step explanation:
To solve the recurrence relation T(n) = 7T(n/2) + 18n^2, we can use iteration or the master theorem.
Let's use iteration:
Let's first write the recurrence relation as:
T(n) = 7T(n/2) + 18n^2.
Next, let's expand T(n/2) in terms of T(n).
After expanding, we get T(n) = 7[7T(n/4) + 18(n/2)^2] + 18n^2.
Continuing this process, we eventually get:
T(n) = 7^k T(n/2^k) + 18n^2 * (1 + 7 + 7^2 + ... + 7^(k-1)).
When n/2^k = 1 (i.e., n = 2^k), we have:
T(n) = 7^k T(1) + 18n^2 * (1 + 7 + 7^2 + ... + 7^(k-1)).
The time complexity can be determined by the value of k.
Since n = 2^k, we can solve for k and obtain k = log2(n).
Therefore, the time complexity of the recurrence relation: T(n) = 7T(n/2) + 18n^2 is O(7^log2(n)).