`(1 +x)^15 = a_0 + a_1x +…..+a_15 x^15 rArr sum_(r = 1)^15 r (a_r)/(a_(r - 1)) = `

Options

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

Answer

To solve the problem, we start with the expression given by the binomial theorem:

\[ (1 + x)^{15} = a_0 + a_1 x + a_2 x^2 + \ldots + a_{15} x^{15} \]

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

We need to find the value of the following summation:

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

### Step 1: Express \( a_r \) and \( a_{r-1} \)

From the binomial theorem, we know:

\[ a_r = \binom{15}{r} \quad \text{and} \quad a_{r-1} = \binom{15}{r-1} \]

### Step 2: Substitute \( a_r \) and \( a_{r-1} \) into the summation

Now we can write:

\[ \frac{a_r}{a_{r-1}} = \frac{\binom{15}{r}}{\binom{15}{r-1}} \]

Using the property of binomial coefficients, we have:

\[ \frac{\binom{15}{r}}{\binom{15}{r-1}} = \frac{15! / (r!(15-r)!)}{(15! / ((r-1)!(15-r+1)!))} = \frac{(15-r+1)}{r} \]

### Step 3: Substitute back into the summation

Thus, we can rewrite the summation as:

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

This simplifies to:

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

### Step 4: Calculate the summation

Now we can compute:

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

The first part is:

\[ \sum_{r=1}^{15} 16 = 16 \times 15 = 240 \]

And the second part is the sum of the first 15 natural numbers:

\[ \sum_{r=1}^{15} r = \frac{15(15 + 1)}{2} = \frac{15 \times 16}{2} = 120 \]

Thus, we have:

\[ \sum_{r=1}^{15} (16 - r) = 240 - 120 = 120 \]

### Final Answer

Therefore, the value of the summation is:

\[ \sum_{r=1}^{15} r \frac{a_r}{a_{r-1}} = 120 \]