Let `A_(n) and B_(n)` be square matrices of order 3, which are defined as :
`A_(n)=[a_(ij)] and B_(n)=[b_(ij)]` where `a_(ij)=(2i+j)/(3^(2n)) and b_(ij)=(3i-j)/(2^(2n))` for all `i` and `j, 1 le i, j le 3`.
If `l=lim_(n to oo) Tr. (3A_(1)+3^(2)A_(2)+3^(3)A_(3)+........+3^(n)A_(n)) and m=lim_(n to oo) Tr. (2B_(1)+2^(2)B_(2)+2^(3)B_(3)+.....+2^(n)B_(n))`, then find the value of `((l+m))/(3)`
[Note : Tr (P) denotes the trace of matrix P.]

Answer

To solve the problem, we need to find the limits \( l \) and \( m \) based on the given matrices \( A_n \) and \( B_n \), and then compute \( \frac{l + m}{3} \).

### Step 1: Define the matrices \( A_n \) and \( B_n \)

Given:
\[ A_{ij} = \frac{2i + j}{3^{2n}} \quad \text{and} \quad B_{ij} = \frac{3i - j}{2^{2n}} \]

For \( n = 1, 2, 3 \), we can write the matrices \( A_n \) and \( B_n \) explicitly.

#### Matrix \( A_n \):
\[ A_n = \frac{1}{3^{2n}} \begin{bmatrix} 2 \cdot 1 + 1 & 2 \cdot 1 + 2 & 2 \cdot 1 + 3 \\ 2 \cdot 2 + 1 & 2 \cdot 2 + 2 & 2 \cdot 2 + 3 \\ 2 \cdot 3 + 1 & 2 \cdot 3 + 2 & 2 \cdot 3 + 3 \end{bmatrix} = \frac{1}{3^{2n}} \begin{bmatrix} 3 & 4 & 5 \\ 5 & 6 & 7 \\ 7 & 8 & 9 \end{bmatrix} \]

#### Matrix \( B_n \):
\[ B_n = \frac{1}{2^{2n}} \begin{bmatrix} 3 \cdot 1 - 1 & 3 \cdot 1 - 2 & 3 \cdot 1 - 3 \\ 3 \cdot 2 - 1 & 3 \cdot 2 - 2 & 3 \cdot 2 - 3 \\ 3 \cdot 3 - 1 & 3 \cdot 3 - 2 & 3 \cdot 3 - 3 \end{bmatrix} = \frac{1}{2^{2n}} \begin{bmatrix} 2 & 1 & 0 \\ 5 & 4 & 3 \\ 8 & 7 & 6 \end{bmatrix} \]

### Step 2: Calculate \( l \)

We need to compute:
\[ l = \lim_{n \to \infty} \text{Tr}(3A_1 + 3^2 A_2 + 3^3 A_3 + \ldots + 3^n A_n) \]

The trace of a matrix is the sum of its diagonal elements. The diagonal elements of \( A_n \) are:
\[ \text{Diagonal of } A_n = \left( \frac{3}{3^{2n}}, \frac{6}{3^{2n}}, \frac{9}{3^{2n}} \right) = \frac{1}{3^{2n}}(3 + 6 + 9) = \frac{18}{3^{2n}} \]

Thus, the trace of \( A_n \):
\[ \text{Tr}(A_n) = \frac{18}{3^{2n}} \]

Now, we compute:
\[ l = \lim_{n \to \infty} \text{Tr}(3A_1 + 3^2 A_2 + \ldots + 3^n A_n) \]
\[ = \lim_{n \to \infty} \left( 3 \cdot \frac{18}{3^{2 \cdot 1}} + 3^2 \cdot \frac{18}{3^{2 \cdot 2}} + \ldots + 3^n \cdot \frac{18}{3^{2n}} \right) \]
\[ = \lim_{n \to \infty} 18 \left( \frac{3}{9} + \frac{9}{27} + \ldots + \frac{3^n}{3^{2n}} \right) \]
This is a geometric series with the first term \( \frac{1}{3} \) and ratio \( \frac{1}{3} \):
\[ = 18 \cdot \frac{\frac{1}{3}}{1 - \frac{1}{3}} = 18 \cdot \frac{1/3}{2/3} = 18 \cdot \frac{1}{2} = 9 \]

### Step 3: Calculate \( m \)

Now we compute:
\[ m = \lim_{n \to \infty} \text{Tr}(2B_1 + 2^2 B_2 + 2^3 B_3 + \ldots + 2^n B_n) \]

The diagonal elements of \( B_n \) are:
\[ \text{Diagonal of } B_n = \left( \frac{2}{2^{2n}}, \frac{4}{2^{2n}}, \frac{6}{2^{2n}} \right) = \frac{12}{2^{2n}} \]

Thus, the trace of \( B_n \):
\[ \text{Tr}(B_n) = \frac{12}{2^{2n}} \]

Now, we compute:
\[ m = \lim_{n \to \infty} \text{Tr}(2B_1 + 2^2 B_2 + \ldots + 2^n B_n) \]
\[ = \lim_{n \to \infty} \left( 2 \cdot \frac{12}{4} + 2^2 \cdot \frac{12}{16} + \ldots + 2^n \cdot \frac{12}{4^n} \right) \]
\[ = \lim_{n \to \infty} 12 \left( \frac{2}{4} + \frac{4}{16} + \ldots + \frac{2^n}{4^n} \right) \]
This is a geometric series with the first term \( \frac{1}{2} \) and ratio \( \frac{1}{2} \):
\[ = 12 \cdot \frac{\frac{1}{2}}{1 - \frac{1}{2}} = 12 \cdot 1 = 12 \]

### Step 4: Compute \( \frac{l + m}{3} \)

Now we have \( l = 9 \) and \( m = 12 \):
\[ \frac{l + m}{3} = \frac{9 + 12}{3} = \frac{21}{3} = 7 \]

### Final Answer
\[ \boxed{7} \]