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

Options

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

Answer

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

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

### Step 1: Simplifying the Binomial Coefficient
We start by simplifying the binomial coefficient:

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

This can be rewritten as:

\[ \frac{{^{15}C_r}}{{^{15}C_{r-1}}} = \frac{{15!}}{{r!(15-r)!}} \cdot \frac{{(16-r)!}}{{(r-1)!}} = \frac{{15! \cdot (16-r)!}}{{r! \cdot (15-r)! \cdot (r-1)!}} \]

### Step 2: Canceling Factorials
By simplifying, we can cancel out the factorials:

\[ = \frac{{16-r}}{{r}} \quad \text{(after simplification)} \]

### Step 3: Substitute Back into the Summation
Now we substitute this back into our summation:

\[ \sum_{r=1}^{15} r^2 \cdot \frac{{16-r}}{r} \]

This simplifies to:

\[ \sum_{r=1}^{15} r(16-r) \]

### Step 4: Splitting the Summation
We can split this summation into two parts:

\[ \sum_{r=1}^{15} (16r - r^2) = 16\sum_{r=1}^{15} r - \sum_{r=1}^{15} r^2 \]

### Step 5: Using Summation Formulas
Now we apply the formulas for the summations:

1. The formula for the sum of the first n natural numbers is:

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

2. The formula for the sum of the squares of the first n natural numbers is:

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

For \( n = 15 \):

\[ \sum_{r=1}^{15} r = \frac{15 \cdot 16}{2} = 120 \]

\[ \sum_{r=1}^{15} r^2 = \frac{15 \cdot 16 \cdot 31}{6} = 1240 \]

### Step 6: Substitute Values Back
Now substituting these values back into our equation:

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

### Conclusion
Thus, the value of the original summation is:

\[ \boxed{680} \]