If `(1+x)^(n)=C_(0)+C_(1)x+…..+C_(n)x^(n)`, then `(C_(1))/(C_(0))+(2C_(2))/(C_(1))+(3C_(3))/(C_(2))+....+(nC_(n))/(C_(n-1))` is :

Options

A. `(2^(n))/(n!)`
B. `(n+1)^(n)/(n!)`
C. `(n(n-1))/(2)`
D. `(n(n+1))/(2)`

Answer

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

\[ \frac{C_1}{C_0} + \frac{2C_2}{C_1} + \frac{3C_3}{C_2} + \ldots + \frac{nC_n}{C_{n-1}} \]

where \( C_r \) represents the binomial coefficients given by \( C_r = \binom{n}{r} \).

### Step-by-Step Solution:

1. **Understanding the Binomial Coefficients**:
The binomial coefficient \( C_r \) is defined as:
\[ C_r = \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
Therefore, we can express \( C_1, C_2, \ldots, C_n \) in terms of \( n \).

2. **Expressing the Ratios**:
We can rewrite the terms in the expression:
\[ \frac{C_1}{C_0} = \frac{\binom{n}{1}}{\binom{n}{0}} = \frac{n}{1} = n \]
\[ \frac{2C_2}{C_1} = \frac{2 \cdot \binom{n}{2}}{\binom{n}{1}} = \frac{2 \cdot \frac{n(n-1)}{2}}{n} = n - 1 \]
\[ \frac{3C_3}{C_2} = \frac{3 \cdot \binom{n}{3}}{\binom{n}{2}} = \frac{3 \cdot \frac{n(n-1)(n-2)}{6}}{\frac{n(n-1)}{2}} = \frac{3(n-2)}{3} = n - 2 \]
Continuing this pattern, we find:
\[ \frac{kC_k}{C_{k-1}} = n - (k - 1) \]

3. **Summing the Series**:
The entire expression can now be rewritten as:
\[ n + (n - 1) + (n - 2) + \ldots + 1 \]
This is the sum of the first \( n \) natural numbers, which can be calculated using the formula:
\[ S = \frac{n(n + 1)}{2} \]

4. **Final Result**:
Therefore, the final result of the given expression is:
\[ \frac{n(n + 1)}{2} \]

### Conclusion:
The value of the expression \( \frac{C_1}{C_0} + \frac{2C_2}{C_1} + \frac{3C_3}{C_2} + \ldots + \frac{nC_n}{C_{n-1}} \) is:

\[ \frac{n(n + 1)}{2} \]