Let `(1 + x)^(n) = sum_(r=0)^(n) C_(r) x^(r)` and ,
`(C_(1))/(C_(0)) + 2 (C_(2))/(C_(1)) + (C_(3))/(C_(2)) +…+ n (C_(n))/(C_(n-1)) = (1)/(k) n(n+1)`,
then the value of k, is

Options

A. `1//2`
B. 2
C. `1//3`
D. 3

Answer

To solve the problem, we need to evaluate the expression given and find the value of \( k \).

### Step-by-Step Solution:

1. **Understanding the Binomial Coefficient**:
We start with the binomial expansion of \( (1 + x)^n \):
\[ (1 + x)^n = \sum_{r=0}^{n} C(n, r) x^r \]
where \( C(n, r) \) is the binomial coefficient, defined as \( C(n, r) = \frac{n!}{r!(n-r)!} \).

2. **Rewriting the Given Expression**:
We need to evaluate the expression:
\[ S_n = \frac{C(1)}{C(0)} + 2 \frac{C(2)}{C(1)} + \frac{C(3)}{C(2)} + \ldots + n \frac{C(n)}{C(n-1)} \]

3. **Using the Property of Binomial Coefficients**:
We know that:
\[ \frac{C(r)}{C(r-1)} = \frac{n - r + 1}{r} \]
Therefore, we can rewrite the expression \( S_n \):
\[ S_n = \sum_{r=1}^{n} r \cdot \frac{C(n, r)}{C(n, r-1)} = \sum_{r=1}^{n} r \cdot \frac{n - r + 1}{r} \]
This simplifies to:
\[ S_n = \sum_{r=1}^{n} (n - r + 1) \]

4. **Calculating the Summation**:
The summation can be split:
\[ S_n = \sum_{r=1}^{n} n - \sum_{r=1}^{n} r + \sum_{r=1}^{n} 1 \]
The first term gives \( n^2 \) (since we have \( n \) added \( n \) times), the second term gives \( \frac{n(n + 1)}{2} \) (the sum of the first \( n \) natural numbers), and the last term gives \( n \):
\[ S_n = n^2 - \frac{n(n + 1)}{2} + n \]

5. **Combining the Terms**:
Combining these terms:
\[ S_n = n^2 + n - \frac{n(n + 1)}{2} \]
To simplify, we can write:
\[ S_n = n^2 + n - \frac{n^2 + n}{2} = n^2 + n - \frac{n^2}{2} - \frac{n}{2} \]
\[ S_n = \frac{2n^2 + 2n - n^2 - n}{2} = \frac{n^2 + n}{2} \]

6. **Setting the Expression Equal to the Given Format**:
We have:
\[ S_n = \frac{n(n + 1)}{2} \]
The problem states:
\[ S_n = \frac{1}{k} n(n + 1) \]
Therefore, equating both expressions gives:
\[ \frac{n(n + 1)}{2} = \frac{1}{k} n(n + 1) \]

7. **Solving for \( k \)**:
Assuming \( n(n + 1) \neq 0 \), we can cancel \( n(n + 1) \) from both sides:
\[ \frac{1}{2} = \frac{1}{k} \]
Thus, we find:
\[ k = 2 \]

### Final Answer:
The value of \( k \) is \( 2 \).