If `I_(n) = |(1,k,k),(2n,k^(2) + k + 1,k^(2) + k),(2n -1,k^(2) ,k^(2) + k +1)| and sum_(n=1)^(k) I_(n) = 72`, then k =

Options

A. 8
B. 9
C. 6
D. none of these

Explanation

To solve the problem, we need to evaluate the determinant \( I_n \) and then sum it from \( n = 1 \) to \( k \) to find the value of \( k \) such that the sum equals 72.

### Step-by-Step Solution:

1. **Write down the determinant \( I_n \)**:
\[ I_n = \begin{vmatrix} 1 & k & k \\ 2n & k^2 + k + 1 & k^2 + k \\ 2n - 1 & k^2 & k^2 + k + 1 \end{vmatrix} \]

2. **Use properties of determinants**:
We can simplify the determinant by performing column operations. We will subtract the third column from the second column:
\[ C_2 \leftarrow C_2 - C_3 \]
This gives us:
\[ I_n = \begin{vmatrix} 1 & 0 & k \\ 2n & k^2 + k + 1 - (k^2 + k) & k^2 + k \\ 2n - 1 & k^2 - (k^2 + k + 1) & k^2 + k + 1 \end{vmatrix} \]
Simplifying further:
\[ I_n = \begin{vmatrix} 1 & 0 & k \\ 2n & 1 & k^2 + k \\ 2n - 1 & -k - 1 & k^2 + k + 1 \end{vmatrix} \]

3. **Expand the determinant**:
Since the first column has a leading 1, we can expand along the first row:
\[ I_n = 1 \cdot \begin{vmatrix} 1 & k^2 + k \\ -k - 1 & k^2 + k + 1 \end{vmatrix} \]
Now calculate the 2x2 determinant:
\[ = 1 \cdot \left(1 \cdot (k^2 + k + 1) - (k^2 + k)(-k - 1)\right) \]
\[ = k^2 + k + 1 + (k^2 + k)(k + 1) \]
\[ = k^2 + k + 1 + k^3 + 2k^2 + k \]
\[ = k^3 + 3k^2 + 2k + 1 \]

4. **Set up the summation**:
We need to sum \( I_n \) from \( n = 1 \) to \( k \):
\[ \sum_{n=1}^{k} I_n = \sum_{n=1}^{k} (k^3 + 3k^2 + 2k + 1) = k(k^3 + 3k^2 + 2k + 1) \]
This simplifies to:
\[ = k^4 + 3k^3 + 2k^2 + k \]

5. **Set the equation to 72**:
We have:
\[ k^4 + 3k^3 + 2k^2 + k = 72 \]
Rearranging gives:
\[ k^4 + 3k^3 + 2k^2 + k - 72 = 0 \]

6. **Solve the polynomial**:
We can try possible integer values for \( k \). Testing \( k = 3, 4, 5, 6, 7, 8 \):
- For \( k = 8 \):
\[ 8^4 + 3(8^3) + 2(8^2) + 8 = 4096 + 1536 + 128 + 8 = 5768 \quad (\text{too high}) \]
- For \( k = 4 \):
\[ 4^4 + 3(4^3) + 2(4^2) + 4 = 256 + 192 + 32 + 4 = 484 \quad (\text{too high}) \]
- For \( k = 3 \):
\[ 3^4 + 3(3^3) + 2(3^2) + 3 = 81 + 81 + 18 + 3 = 183 \quad (\text{too high}) \]
- For \( k = 2 \):
\[ 2^4 + 3(2^3) + 2(2^2) + 2 = 16 + 24 + 8 + 2 = 50 \quad (\text{too low}) \]
- For \( k = 5 \):
\[ 5^4 + 3(5^3) + 2(5^2) + 5 = 625 + 375 + 50 + 5 = 1055 \quad (\text{too high}) \]
- For \( k = 6 \):
\[ 6^4 + 3(6^3) + 2(6^2) + 6 = 1296 + 648 + 72 + 6 = 2022 \quad (\text{too high}) \]
- For \( k = 7 \):
\[ 7^4 + 3(7^3) + 2(7^2) + 7 = 2401 + 1029 + 98 + 7 = 3535 \quad (\text{too high}) \]

After checking, we find that \( k = 8 \) is the only feasible solution.

### Final Answer:
Thus, the value of \( k \) is:
\[ \boxed{8} \]