If `C_(0),C_(1),C_(2),… C_(15)` are the binomial coefficients in the expansion of `(1+x)^(15)`, then
`(C_(1))/(C_(0)) +2(C_(2))/(C_(1)) +3(C_(3))/(C_(2))+…+15(C_(15))/(C_(14))=`

Options

A. 120
B. 130
C. 140
D. 150

Answer

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

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

where \( C_n \) are the binomial coefficients from the expansion of \( (1+x)^{15} \).

### Step-by-Step Solution:

1. **Understanding Binomial Coefficients**:
The binomial coefficients \( C_n \) can be defined as:
\[ C_n = \binom{15}{n} = \frac{15!}{n!(15-n)!} \]
Therefore, we have:
- \( C_0 = \binom{15}{0} = 1 \)
- \( C_1 = \binom{15}{1} = 15 \)
- \( C_2 = \binom{15}{2} = \frac{15 \cdot 14}{2} = 105 \)
- and so on, up to \( C_{15} = \binom{15}{15} = 1 \).

2. **Expressing the Terms**:
We can express the terms in the sum:
\[ \frac{C_1}{C_0} = \frac{15}{1} = 15 \]
\[ \frac{C_2}{C_1} = \frac{105}{15} = 7 \]
\[ \frac{C_3}{C_2} = \frac{455}{105} = \frac{91}{21} = \frac{13}{3} \]
Continuing this way, we can see that each term can be expressed as:
\[ n \cdot \frac{C_n}{C_{n-1}} = n \cdot \frac{\binom{15}{n}}{\binom{15}{n-1}} = n \cdot \frac{15-n+1}{n} = 15 - n + 1 = 16 - n \]

3. **Summing the Series**:
The entire expression can be simplified as:
\[ \sum_{n=1}^{15} (16 - n) = 16 \cdot 15 - \sum_{n=1}^{15} n \]
The sum \( \sum_{n=1}^{15} n \) can be calculated using the formula for the sum of the first \( n \) natural numbers:
\[ \sum_{n=1}^{k} n = \frac{k(k+1)}{2} \]
For \( k = 15 \):
\[ \sum_{n=1}^{15} n = \frac{15 \cdot 16}{2} = 120 \]
Therefore, we have:
\[ 16 \cdot 15 - 120 = 240 - 120 = 120 \]

4. **Final Result**:
Thus, the value of the expression is:
\[ \frac{C_1}{C_0} + 2 \cdot \frac{C_2}{C_1} + 3 \cdot \frac{C_3}{C_2} + \ldots + 15 \cdot \frac{C_{15}}{C_{14}} = 120 \]