To determine the worst-case time complexity of inserting $ n^2 $ elements into an AVL tree that initially contains $ n $ elements, we'll analyze the time it takes to insert each element and sum these times.

Understanding AVL Trees:

AVL Tree Properties:

Height-Balanced Binary Search Tree: Ensures that the heights of the two child subtrees of any node differ by at most one.

Operations Time Complexity:

Insertion: $ O(\log k) $, where $ k $ is the number of nodes in the tree at the time of insertion.

Balancing: After insertion, rebalancing the tree requires at most $ O(1) $ rotations, which is $ O(1) $ time.

Analyzing the Insertion of $ n^2 $ Elements:

Initial Tree Size: $ n $

Total Number of Insertions: $ n^2 $

Tree Size After Insertions: $ n + n^2 = n^2 + n $

Time to Insert Each Element:

The time to insert the $ i^{th} $ element is $ O(\log(N_i)) $, where $ N_i $ is the size of the tree before the $ i^{th} $ insertion.

Since the tree grows by one with each insertion, $ N_i = n + i - 1 $.

Total Time Complexity ($ T $):

$ T = \sum_{i=1}^{n^2} O\left( \log (n + i - 1) \right) $

Calculating the Total Time Complexity:

Step 1: Estimate the Range of $ \log (n + i - 1) $

Minimum Value:

When $ i = 1 $:

$ \log (n + 1 - 1) = \log n $

Maximum Value:

When $ i = n^2 $:

$ \log (n + n^2 - 1) \approx \log n^2 = 2 \log n $

Step 2: Approximate the Sum

The logarithm function increases slowly, so over the range from $ n $ to $ n^2 + n - 1 $, $ \log (n + i - 1) $ is between $ \log n $ and $ 2 \log n $.

Average Value of $ \log (n + i - 1) $:

Approximately $ \frac{1}{2} ( \log n + 2 \log n ) = \frac{3}{2} \log n $

Step 3: Compute the Total Time

Total Time:

$ T = n^2 \times O(\log n) = O(n^2 \log n) $

Conclusion:

The total worst-case time complexity for inserting $ n^2 $ elements into an AVL tree that initially contains $ n $ elements is $ \Theta(n^2 \log n) $.

This is because each insertion takes $ O(\log k) $ time, where $ k $ is the size of the tree at that point, and we sum this over all $ n^2 $ insertions.

Answer: Option B

$\Theta\left(n^2 \log n\right)$