If `a_0, a_1 , a_2 …..a_n` are binomial coefficients then `(1 + a_1/a_0)(1 +a_2/a_1) …………….(1 + (a_n)/(a_(n-1)) ) = `

Options

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

Answer

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

\[ (1 + \frac{a_1}{a_0})(1 + \frac{a_2}{a_1}) \ldots (1 + \frac{a_n}{a_{n-1}}) \]

where \( a_k \) are the binomial coefficients \( \binom{n}{k} \).

### Step-by-Step Solution:

1. **Identify the Binomial Coefficients**:
The binomial coefficients are defined as:
\[ a_k = \binom{n}{k} \]
for \( k = 0, 1, 2, \ldots, n \).

2. **Rewrite the Expression**:
Substitute the binomial coefficients into the expression:
\[ (1 + \frac{\binom{n}{1}}{\binom{n}{0}})(1 + \frac{\binom{n}{2}}{\binom{n}{1}}) \ldots (1 + \frac{\binom{n}{n}}{\binom{n}{n-1}}) \]

3. **Simplify Each Term**:
Each term can be simplified:
\[ 1 + \frac{\binom{n}{k}}{\binom{n}{k-1}} = 1 + \frac{n!}{k!(n-k)!} \cdot \frac{(k-1)!(n-k+1)!}{n!} = 1 + \frac{n-k+1}{k} = \frac{n-k+1 + k}{k} = \frac{n+1-k}{k} \]

4. **Combine the Terms**:
Now, we can write the entire product:
\[ \prod_{k=1}^{n} \left(1 + \frac{a_k}{a_{k-1}}\right) = \prod_{k=1}^{n} \frac{n+1-k}{k} \]

5. **Recognize the Pattern**:
The product can be rewritten as:
\[ \frac{(n+1-1)(n+1-2)\ldots(n+1-n)}{1 \cdot 2 \cdot \ldots \cdot n} = \frac{n!}{n!} = 1 \]

6. **Final Result**:
Thus, the final result of the expression is:
\[ (1 + \frac{a_1}{a_0})(1 + \frac{a_2}{a_1}) \ldots (1 + \frac{a_n}{a_{n-1}}) = \frac{(n+1)!}{n!} = n + 1 \]

### Conclusion:
The value of the expression is \( n + 1 \).