Find the sum of the following
`(""^(15)C_(1))/(""^(15)C_(0))+2(""^(15)C_(2))/(""^(15)C_(1))+3(""^(15)C_(3))/(""^(15)C_(2))+......+15(""^(15)C_(15))/(""^(15)C_(14))`

Options

A. `2^15`
B. 240
C. 120
D. 136

Answer

To solve the problem, we need to find the sum of the following expression:

\[ \frac{1 \cdot {^{15}C_{1}}}{^{15}C_{0}} + 2 \cdot \frac{^{15}C_{2}}{^{15}C_{1}} + 3 \cdot \frac{^{15}C_{3}}{^{15}C_{2}} + \ldots + 15 \cdot \frac{^{15}C_{15}}{^{15}C_{14}} \]

### Step 1: Understanding the terms in the sum

The general term in the sum can be expressed as:

\[ k \cdot \frac{^{15}C_k}{^{15}C_{k-1}} \quad \text{for } k = 1, 2, \ldots, 15 \]

Using the property of binomial coefficients, we know that:

\[ \frac{^{n}C_{k}}{^{n}C_{k-1}} = \frac{n-k+1}{k} \]

### Step 2: Simplifying each term

For our case where \( n = 15 \):

\[ \frac{^{15}C_k}{^{15}C_{k-1}} = \frac{15 - k + 1}{k} = \frac{16 - k}{k} \]

Thus, the general term becomes:

\[ k \cdot \frac{^{15}C_k}{^{15}C_{k-1}} = k \cdot \frac{16 - k}{k} = 16 - k \]

### Step 3: Writing the entire sum

Now, we can rewrite the entire sum:

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

### Step 4: Evaluating the sum

We can separate the sum:

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

Calculating each part:

1. The first part:

\[ \sum_{k=1}^{15} 16 = 16 \cdot 15 = 240 \]

2. The second part (using the formula for the sum of the first \( n \) natural numbers):

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

### Step 5: Final calculation

Now substituting back into our sum:

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

Thus, the final result is:

\[ \boxed{120} \]