Let `(1 + x)^(n) = sum_(r=0)^(n) a_(r) x^(r)` . Then
`( a+ (a_(1))/(a_(0))) (1 + (a_(2))/(a_(1)))…(1 + (a_(n))/(a_(n-1)))` is equal to

Options

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

Answer

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

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

where \(a_r\) are the coefficients from the binomial expansion of \((1 + x)^n\).

### Step 1: Identify the coefficients \(a_r\)
From the binomial theorem, we know that:

\[ (1 + x)^n = \sum_{r=0}^{n} a_r x^r \]

where \(a_r = \binom{n}{r}\). Thus, we have:

- \(a_0 = \binom{n}{0} = 1\)
- \(a_1 = \binom{n}{1} = n\)
- \(a_2 = \binom{n}{2} = \frac{n(n-1)}{2}\)
- \(\ldots\)
- \(a_n = \binom{n}{n} = 1\)

### Step 2: Write the expression in terms of \(n\)
Now, we can express each term in the product:

\[ 1 + \frac{a_r}{a_{r-1}} = 1 + \frac{\binom{n}{r}}{\binom{n}{r-1}} = 1 + \frac{n - (r - 1)}{r} = 1 + \frac{n - r + 1}{r} \]

### Step 3: Simplify each term
The term simplifies to:

\[ 1 + \frac{n - r + 1}{r} = \frac{r + n - r + 1}{r} = \frac{n + 1}{r} \]

### Step 4: Write the full product
Now we can write the full product:

\[ \prod_{r=1}^{n} \left(1 + \frac{a_r}{a_{r-1}}\right) = \prod_{r=1}^{n} \frac{n + 1}{r} \]

### Step 5: Evaluate the product
This can be simplified as:

\[ = (n + 1)^n \prod_{r=1}^{n} \frac{1}{r} = \frac{(n + 1)^n}{n!} \]

### Final Result
Thus, the final result is:

\[ \frac{(n + 1)^n}{n!} \]