If `(1+x) ^(15) =a_(0) +a_(1) x +a_(2) x ^(2) +…+ a_(15) x ^(15),` then the value of `sum_(r=1) ^(15) r . (a_(r))/(a _(r-1))` is-

Options

A. 110
B. 115
C. 120
D. 135

Answer

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

\[ \sum_{r=1}^{15} r \cdot \frac{a_r}{a_{r-1}} \]

where \( a_r = \binom{15}{r} \) and \( a_{r-1} = \binom{15}{r-1} \).

### Step-by-Step Solution:

1. **Identify the coefficients**: From the binomial expansion of \( (1+x)^{15} \), we know:
\[ a_r = \binom{15}{r} \]
and
\[ a_{r-1} = \binom{15}{r-1} \]

2. **Express the ratio**: We can express the ratio \( \frac{a_r}{a_{r-1}} \):
\[ \frac{a_r}{a_{r-1}} = \frac{\binom{15}{r}}{\binom{15}{r-1}} = \frac{15! / (r!(15-r)!)}{15! / ((r-1)!(15-r+1)!)} = \frac{(15-r+1)}{r} \]

3. **Substitute the ratio into the summation**: Now we substitute this back into our summation:
\[ \sum_{r=1}^{15} r \cdot \frac{a_r}{a_{r-1}} = \sum_{r=1}^{15} r \cdot \frac{15-r+1}{r} \]
This simplifies to:
\[ \sum_{r=1}^{15} (15 - r + 1) = \sum_{r=1}^{15} (16 - r) \]

4. **Change the index of summation**: We can rewrite the summation:
\[ \sum_{r=1}^{15} (16 - r) = \sum_{r=1}^{15} 16 - \sum_{r=1}^{15} r \]
The first part is simply \( 16 \times 15 \) and the second part is the sum of the first 15 natural numbers.

5. **Calculate the sums**:
- The sum of the first 15 natural numbers is given by:
\[ \sum_{r=1}^{15} r = \frac{15 \cdot (15 + 1)}{2} = \frac{15 \cdot 16}{2} = 120 \]
- Therefore, we have:
\[ \sum_{r=1}^{15} (16 - r) = 16 \cdot 15 - 120 = 240 - 120 = 120 \]

### Final Answer:
Thus, the value of \( \sum_{r=1}^{15} r \cdot \frac{a_r}{a_{r-1}} \) is:

\[ \boxed{120} \]