The value of `sum_(r=0)^(15)r^(2)((""^(15)C_(r))/(""^(15)C_(r-1)))` is equal to

Options

A. 1085
B. 560
C. 680
D. 1240

Explanation

To solve the problem, we need to evaluate the sum:

\[ \sum_{r=0}^{15} r^2 \left( \frac{15C_r}{15C_{r-1}} \right) \]

### Step 1: Simplifying the Binomial Coefficient Ratio

We start by simplifying the ratio of the binomial coefficients:

\[ \frac{15C_r}{15C_{r-1}} = \frac{\frac{15!}{r!(15-r)!}}{\frac{15!}{(r-1)!(15-r+1)!}} = \frac{(r-1)!(15-r+1)!}{r!(15-r)!} \]

This simplifies to:

\[ \frac{15-r+1}{r} = \frac{16-r}{r} \]

### Step 2: Substituting Back into the Sum

Now we substitute this back into the original sum:

\[ \sum_{r=0}^{15} r^2 \cdot \frac{16 - r}{r} = \sum_{r=0}^{15} (16r - r^2) \]

### Step 3: Splitting the Sum

We can split the sum into two separate sums:

\[ \sum_{r=0}^{15} (16r - r^2) = 16 \sum_{r=0}^{15} r - \sum_{r=0}^{15} r^2 \]

### Step 4: Calculating the Sums

1. **Calculating \(\sum_{r=0}^{15} r\)**:
The formula for the sum of the first \(n\) natural numbers is:

\[ \sum_{r=0}^{n} r = \frac{n(n+1)}{2} \]

For \(n = 15\):

\[ \sum_{r=0}^{15} r = \frac{15 \cdot 16}{2} = 120 \]

2. **Calculating \(\sum_{r=0}^{15} r^2\)**:
The formula for the sum of the squares of the first \(n\) natural numbers is:

\[ \sum_{r=0}^{n} r^2 = \frac{n(n+1)(2n+1)}{6} \]

For \(n = 15\):

\[ \sum_{r=0}^{15} r^2 = \frac{15 \cdot 16 \cdot 31}{6} = 1240 \]

### Step 5: Substituting Back into the Expression

Now substituting these values back into our expression:

\[ 16 \cdot 120 - 1240 = 1920 - 1240 = 680 \]

### Final Answer

Thus, the value of the sum is:

\[ \boxed{680} \]