Let `C_r `= `""^(15)C_(r), 0ler <=15`. Sum of the series `S=sum_(r=1)^(15)r(C_(r))/(C_(r-1))` is

Options

A. 40
B. 60
C. 100
D. 120

Answer

To solve the problem, we need to find the sum of the series given by:

\[ S = \sum_{r=1}^{15} r \frac{C_r}{C_{r-1}} \]

where \( C_r = \binom{15}{r} \).

### Step-by-Step Solution:

1. **Understanding the Binomial Coefficient**:
The binomial coefficient \( C_r \) can be expressed as:
\[ C_r = \binom{15}{r} = \frac{15!}{r!(15-r)!} \]
and
\[ C_{r-1} = \binom{15}{r-1} = \frac{15!}{(r-1)!(15-(r-1))!} = \frac{15!}{(r-1)!(16-r)!} \]

2. **Expressing the Ratio**:
We need to compute the ratio \( \frac{C_r}{C_{r-1}} \):
\[ \frac{C_r}{C_{r-1}} = \frac{\frac{15!}{r!(15-r)!}}{\frac{15!}{(r-1)!(16-r)!}} = \frac{(16-r)!}{(15-r)!} \cdot \frac{1}{r!} \cdot (r-1)! = \frac{16-r}{r} \]

3. **Substituting Back into the Series**:
Now substituting this back into our series:
\[ S = \sum_{r=1}^{15} r \cdot \frac{16-r}{r} = \sum_{r=1}^{15} (16 - r) \]

4. **Simplifying the Series**:
The series can be simplified:
\[ S = \sum_{r=1}^{15} (16 - r) = \sum_{r=1}^{15} 16 - \sum_{r=1}^{15} r \]
The first part is:
\[ \sum_{r=1}^{15} 16 = 16 \times 15 = 240 \]
The second part is the sum of the first 15 natural numbers:
\[ \sum_{r=1}^{15} r = \frac{15 \times 16}{2} = 120 \]

5. **Calculating the Final Sum**:
Now substituting back:
\[ S = 240 - 120 = 120 \]

Thus, the sum of the series \( S \) is:

\[ \boxed{120} \]