If `C_(r )=""^(n)C_(r )`, then the value of
`2((C_(1))/(C_(0)) +2(C_(2))/(C_(1)) +3(C_(3))/(C_(2)) +…+n.(C_(n))/(C_(n-1)))` is

Options

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

Answer

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

\[ 2\left(\frac{C_1}{C_0} + 2\frac{C_2}{C_1} + 3\frac{C_3}{C_2} + \ldots + n\frac{C_n}{C_{n-1}}\right) \]

where \(C_r\) denotes the binomial coefficient \(\binom{n}{r}\).

### Step-by-Step Solution:

1. **Identify the General Term**:
The general term in the series can be represented as:
\[ T_r = r \cdot \frac{C_r}{C_{r-1}} \]
where \(C_r = \binom{n}{r}\) and \(C_{r-1} = \binom{n}{r-1}\).

2. **Express the Binomial Coefficients**:
Using the formula for binomial coefficients:
\[ C_r = \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
and
\[ C_{r-1} = \binom{n}{r-1} = \frac{n!}{(r-1)!(n-r+1)!} \]

3. **Substituting the Coefficients**:
Substitute these into the general term \(T_r\):
\[ T_r = r \cdot \frac{\binom{n}{r}}{\binom{n}{r-1}} = r \cdot \frac{\frac{n!}{r!(n-r)!}}{\frac{n!}{(r-1)!(n-r+1)!}} = r \cdot \frac{(n-r+1)}{r} = n - r + 1 \]

4. **Summing the Series**:
Now, we need to sum \(T_r\) from \(r=1\) to \(n\):
\[ S = \sum_{r=1}^{n} T_r = \sum_{r=1}^{n} (n - r + 1) \]
This can be rewritten as:
\[ S = \sum_{r=1}^{n} (n + 1 - r) = \sum_{r=1}^{n} (n + 1) - \sum_{r=1}^{n} r \]
The first sum is simply \((n + 1)n\) and the second sum is \(\frac{n(n + 1)}{2}\):
\[ S = (n + 1)n - \frac{n(n + 1)}{2} = \frac{2(n + 1)n - n(n + 1)}{2} = \frac{n(n + 1)}{2} \]

5. **Final Calculation**:
Now, we multiply \(S\) by 2 as per the original expression:
\[ 2S = 2 \cdot \frac{n(n + 1)}{2} = n(n + 1) \]

Thus, the value of the given expression is:

\[ \boxed{n(n + 1)} \]