`A=[(1,1,1),(0,1,1),(0,0,1)]` , then sum of all elements in `A+A^2+A^3+ . . . + A^(26)` is

Answer

To solve the problem, we need to find the sum of all elements in the series \( A + A^2 + A^3 + \ldots + A^{26} \), where the matrix \( A \) is given as:

\[ A = \begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix} \]

### Step 1: Calculate \( A^2 \)

To find \( A^2 \), we multiply \( A \) by itself:

\[ A^2 = A \cdot A = \begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix} \]

Calculating the elements of \( A^2 \):

- First row:
- \( (1 \cdot 1 + 1 \cdot 0 + 1 \cdot 0) = 1 \)
- \( (1 \cdot 1 + 1 \cdot 1 + 1 \cdot 0) = 2 \)
- \( (1 \cdot 1 + 1 \cdot 1 + 1 \cdot 1) = 3 \)

- Second row:
- \( (0 \cdot 1 + 1 \cdot 0 + 1 \cdot 0) = 0 \)
- \( (0 \cdot 1 + 1 \cdot 1 + 1 \cdot 0) = 1 \)
- \( (0 \cdot 1 + 1 \cdot 1 + 1 \cdot 1) = 2 \)

- Third row:
- \( (0 \cdot 1 + 0 \cdot 0 + 1 \cdot 0) = 0 \)
- \( (0 \cdot 1 + 0 \cdot 1 + 1 \cdot 0) = 0 \)
- \( (0 \cdot 1 + 0 \cdot 1 + 1 \cdot 1) = 1 \)

Thus,

\[ A^2 = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{pmatrix} \]

### Step 2: Calculate \( A^3 \)

Next, we calculate \( A^3 \):

\[ A^3 = A^2 \cdot A = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix} \]

Calculating the elements of \( A^3 \):

- First row:
- \( (1 \cdot 1 + 2 \cdot 0 + 3 \cdot 0) = 1 \)
- \( (1 \cdot 1 + 2 \cdot 1 + 3 \cdot 0) = 3 \)
- \( (1 \cdot 1 + 2 \cdot 1 + 3 \cdot 1) = 6 \)

- Second row:
- \( (0 \cdot 1 + 1 \cdot 0 + 2 \cdot 0) = 0 \)
- \( (0 \cdot 1 + 1 \cdot 1 + 2 \cdot 0) = 1 \)
- \( (0 \cdot 1 + 1 \cdot 1 + 2 \cdot 1) = 3 \)

- Third row:
- \( (0 \cdot 1 + 0 \cdot 0 + 1 \cdot 0) = 0 \)
- \( (0 \cdot 1 + 0 \cdot 1 + 1 \cdot 0) = 0 \)
- \( (0 \cdot 1 + 0 \cdot 1 + 1 \cdot 1) = 1 \)

Thus,

\[ A^3 = \begin{pmatrix} 1 & 3 & 6 \\ 0 & 1 & 3 \\ 0 & 0 & 1 \end{pmatrix} \]

### Step 3: Identify the Pattern

From the calculations, we can observe a pattern in the matrices:

\[ A^n = \begin{pmatrix} 1 & n & \frac{n(n+1)}{2} \\ 0 & 1 & n \\ 0 & 0 & 1 \end{pmatrix} \]

### Step 4: Calculate \( A^{26} \)

Using the pattern, we can write:

\[ A^{26} = \begin{pmatrix} 1 & 26 & \frac{26 \cdot 27}{2} \\ 0 & 1 & 26 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 26 & 351 \\ 0 & 1 & 26 \\ 0 & 0 & 1 \end{pmatrix} \]

### Step 5: Sum of All Elements

Now, we need to find the sum of all elements in \( A + A^2 + A^3 + \ldots + A^{26} \).

The sum of elements in \( A^n \) can be calculated as follows:

- The sum of elements in \( A^n \) is \( 1 + n + \frac{n(n+1)}{2} + 0 + 1 + n + 0 + 0 + 1 = 3 + 2n + \frac{n(n+1)}{2} \).

Thus, the total sum can be computed as:

\[ \text{Total Sum} = \sum_{n=1}^{26} \left( 3 + 2n + \frac{n(n+1)}{2} \right) \]

Calculating each part:

1. **Sum of constants**: \( 3 \times 26 = 78 \)
2. **Sum of \( 2n \)**: \( 2 \times \frac{26 \cdot 27}{2} = 26 \cdot 27 = 702 \)
3. **Sum of \( \frac{n(n+1)}{2} \)**:
- \( \sum_{n=1}^{26} \frac{n(n+1)}{2} = \frac{1}{2} \sum_{n=1}^{26} n^2 + \frac{1}{2} \sum_{n=1}^{26} n \)
- \( = \frac{1}{2} \cdot \frac{26 \cdot 27 \cdot 53}{6} + \frac{1}{2} \cdot \frac{26 \cdot 27}{2} = 6201 + 351 = 6552 \)

Adding these sums together:

\[ \text{Total Sum} = 78 + 702 + 6552 = 6332 \]

### Final Answer

Thus, the sum of all elements in \( A + A^2 + A^3 + \ldots + A^{26} \) is:

\[ \boxed{4056} \]