Let `C_(r)=""^(15)C_(r),(0lerle15)`, and `m=(C_(1))/(C_(0))+(2C_(3))/(C_(2))+(3C_(3))/(C_(2))+….+(""^(15)C_(15))/(C_(14))`
then m = _______

Answer

To solve the problem, we need to evaluate the expression for \( m \):

\[ m = \frac{C_1}{C_0} + \frac{2C_3}{C_2} + \frac{3C_3}{C_2} + \ldots + \frac{C_{15}}{C_{14}} \]

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

### Step 1: Rewrite the terms

We can rewrite each term in the sum using the binomial coefficient notation:

\[ m = \sum_{r=1}^{15} \frac{rC_r}{C_{r-1}} \]

### Step 2: Simplify the ratio of binomial coefficients

Using the property of binomial coefficients, we know that:

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

Now, we can express \( \frac{C_r}{C_{r-1}} \):

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

### Step 3: Substitute back into the sum

Now substituting back into our expression for \( m \):

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

### Step 4: Evaluate the sum

The sum \( \sum_{r=1}^{15} (16 - r) \) can be simplified:

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

The first part is simply \( 16 \times 15 \) (since there are 15 terms):

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

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

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

### Step 5: Combine the results

Now, substituting back into the equation for \( m \):

\[ m = 240 - 120 = 120 \]

Thus, the final answer is:

\[ \boxed{120} \]