If `D_r=|{:(r,15,8),(r^2,35,9),(r^3,25,10):}|`, then the value of `root(5)(((-1/100)sum_(r=1)^5D_r)-37)` is equal to

Options

A. 5
B. 2
C. 9
D. 3

Answer

To solve the problem, we need to evaluate the determinant \( D_r = \begin{vmatrix} r & 15 & 8 \\ r^2 & 35 & 9 \\ r^3 & 25 & 10 \end{vmatrix} \) for \( r = 1, 2, 3, 4, 5 \), then compute the expression \( \sqrt[5]{\left(-\frac{1}{100} \sum_{r=1}^{5} D_r\right) - 37} \).

### Step 1: Calculate the determinant \( D_r \)

The determinant \( D_r \) can be calculated using the formula for 3x3 determinants:

\[ D_r = r \begin{vmatrix} 35 & 9 \\ 25 & 10 \end{vmatrix} - 15 \begin{vmatrix} r^2 & 9 \\ r^3 & 10 \end{vmatrix} + 8 \begin{vmatrix} r^2 & 35 \\ r^3 & 25 \end{vmatrix} \]

Calculating the 2x2 determinants:

1. \( \begin{vmatrix} 35 & 9 \\ 25 & 10 \end{vmatrix} = (35 \cdot 10) - (9 \cdot 25) = 350 - 225 = 125 \)

2. \( \begin{vmatrix} r^2 & 9 \\ r^3 & 10 \end{vmatrix} = (r^2 \cdot 10) - (9 \cdot r^3) = 10r^2 - 9r^3 \)

3. \( \begin{vmatrix} r^2 & 35 \\ r^3 & 25 \end{vmatrix} = (r^2 \cdot 25) - (35 \cdot r^3) = 25r^2 - 35r^3 \)

Substituting these back into the determinant expression:

\[ D_r = r \cdot 125 - 15(10r^2 - 9r^3) + 8(25r^2 - 35r^3) \]

Expanding this:

\[ D_r = 125r - 150r^2 + 135r^3 + 200r^2 - 280r^3 \]

Combining like terms:

\[ D_r = 125r + (200 - 150)r^2 + (135 - 280)r^3 = 125r + 50r^2 - 145r^3 \]

### Step 2: Calculate \( \sum_{r=1}^{5} D_r \)

Now we need to compute \( \sum_{r=1}^{5} D_r \):

\[ \sum_{r=1}^{5} D_r = \sum_{r=1}^{5} (125r + 50r^2 - 145r^3) \]

Using the formulas for summation:

1. \( \sum_{r=1}^{n} r = \frac{n(n+1)}{2} \)
2. \( \sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6} \)
3. \( \sum_{r=1}^{n} r^3 = \left(\frac{n(n+1)}{2}\right)^2 \)

For \( n = 5 \):

- \( \sum_{r=1}^{5} r = \frac{5 \cdot 6}{2} = 15 \)
- \( \sum_{r=1}^{5} r^2 = \frac{5 \cdot 6 \cdot 11}{6} = 55 \)
- \( \sum_{r=1}^{5} r^3 = \left(\frac{5 \cdot 6}{2}\right)^2 = 225 \)

Now substituting these values:

\[ \sum_{r=1}^{5} D_r = 125 \cdot 15 + 50 \cdot 55 - 145 \cdot 225 \]

Calculating each term:

1. \( 125 \cdot 15 = 1875 \)
2. \( 50 \cdot 55 = 2750 \)
3. \( -145 \cdot 225 = -32625 \)

Combining these:

\[ \sum_{r=1}^{5} D_r = 1875 + 2750 - 32625 = 4625 - 32625 = -28000 \]

### Step 3: Substitute into the expression

Now we substitute \( \sum_{r=1}^{5} D_r \) into the expression:

\[ \sqrt[5]{\left(-\frac{1}{100} \cdot (-28000)\right) - 37} \]

Calculating:

\[ -\frac{1}{100} \cdot (-28000) = 280 \]

So we have:

\[ \sqrt[5]{280 - 37} = \sqrt[5]{243} \]

Since \( 243 = 3^5 \):

\[ \sqrt[5]{243} = 3 \]

### Final Answer

Thus, the value of the expression is:

\[ \boxed{3} \]