`A_(1)=[a_(1)]`
`A_(2)=[{:(,a_(2),a_(3)),(,a_(4),a_(5)):}]`
`A_(3)=[{:(,a_(6),a_(7),a_(8)),(,a_(9),a_(10),a_(11)),(,a_(12),a_(13),a_(14)):}]......A_(n)=[.......]`
Where, `a_(r)=[log_(2)r]([.])` denotes greatest integer). Then trace of `A_(10)`

Answer

To find the trace of the matrix \( A_{10} \), we will follow these steps:

### Step 1: Understand the structure of the matrices
The matrices are defined as follows:
- \( A_1 = [a_1] \) is a \( 1 \times 1 \) matrix.
- \( A_2 = \begin{bmatrix} a_2 & a_3 \\ a_4 & a_5 \end{bmatrix} \) is a \( 2 \times 2 \) matrix.
- \( A_3 = \begin{bmatrix} a_6 & a_7 & a_8 \\ a_9 & a_{10} & a_{11} \\ a_{12} & a_{13} & a_{14} \end{bmatrix} \) is a \( 3 \times 3 \) matrix.
- Continuing this pattern, \( A_n \) is an \( n \times n \) matrix.

### Step 2: Determine the number of elements in \( A_9 \)
The total number of elements in \( A_9 \) can be calculated using the formula for the sum of squares:
\[ \text{Total elements} = 1^2 + 2^2 + 3^2 + \ldots + 9^2 = \frac{n(n + 1)(2n + 1)}{6} \]
For \( n = 9 \):
\[ \text{Total elements} = \frac{9 \cdot 10 \cdot 19}{6} = 285 \]

### Step 3: Identify the elements of \( A_{10} \)
Since \( A_9 \) covers elements \( a_1 \) to \( a_{285} \), the elements of \( A_{10} \) will be:
\[ a_{286}, a_{287}, \ldots, a_{295} \]
This gives us a \( 10 \times 10 \) matrix \( A_{10} \).

### Step 4: Find the diagonal elements of \( A_{10} \)
The diagonal elements of \( A_{10} \) are:
\[ a_{286}, a_{297}, a_{308}, \ldots, a_{385} \]
These correspond to the indices \( 286, 297, 308, \ldots, 385 \).

### Step 5: Calculate the trace of \( A_{10} \)
The trace of a matrix is the sum of its diagonal elements:
\[ \text{Trace}(A_{10}) = a_{286} + a_{297} + a_{308} + \ldots + a_{385} \]

### Step 6: Evaluate \( a_r \)
Given \( a_r = \lfloor \log_2 r \rfloor \), we need to evaluate:
- For \( r = 286 \): \( \lfloor \log_2 286 \rfloor = 8 \)
- For \( r = 297 \): \( \lfloor \log_2 297 \rfloor = 8 \)
- For \( r = 308 \): \( \lfloor \log_2 308 \rfloor = 8 \)
- Continuing this, we find that all values from \( 286 \) to \( 295 \) yield \( 8 \).

### Step 7: Sum the contributions to the trace
Since there are \( 10 \) diagonal elements, all equal to \( 8 \):
\[ \text{Trace}(A_{10}) = 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 = 80 \]

### Final Answer
The trace of \( A_{10} \) is \( 80 \).

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