If ` (1 + x)^(n) = sum_(r=0)^(n) C_(r) x^(r),(1 + (C_(1))/(C_(0))) (1 + (C_(2))/(C_(1)))...(1 + (C_(n))/(C_(n-1))) ` is equal to

Options

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

Answer

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

\[ (1 + \frac{C_1}{C_0})(1 + \frac{C_2}{C_1})...(1 + \frac{C_n}{C_{n-1}}) \]

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

### Step-by-step Solution:

1. **Understanding the Binomial Coefficients**:
The binomial coefficients are defined as:
\[ C_r = \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
Therefore, we can express the ratios:
\[ \frac{C_r}{C_{r-1}} = \frac{\binom{n}{r}}{\binom{n}{r-1}} = \frac{n!/(r!(n-r)!)}{(n!/(r-1)!(n-r+1)!)} = \frac{(n-r+1)}{r} \]

2. **Substituting the Ratios**:
Now, substituting this into our expression, we have:
\[ 1 + \frac{C_r}{C_{r-1}} = 1 + \frac{(n-r+1)}{r} = \frac{r + (n-r+1)}{r} = \frac{n+1}{r} \]

3. **Writing the Full Product**:
Thus, the entire product becomes:
\[ \prod_{r=1}^{n} \left(1 + \frac{C_r}{C_{r-1}}\right) = \prod_{r=1}^{n} \frac{n+1}{r} \]

4. **Simplifying the Product**:
The product can be simplified as:
\[ \prod_{r=1}^{n} \frac{n+1}{r} = \frac{(n+1)^n}{1 \cdot 2 \cdot 3 \cdots n} = \frac{(n+1)^n}{n!} \]

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

### Final Answer:
\[ \frac{(n+1)^n}{n!} \]