If ` (1 + x)^(n) = sum_(r=0)^(n) C_(r) x^(r),(1 + (C_(0))/(C_(1))) (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:

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

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

### Step 1: Write the binomial coefficients

The binomial coefficients are defined as:

\[ C_r = \binom{n}{r} = \frac{n!}{r!(n-r)!} \]

Thus, we can express \(C_0\), \(C_1\), \(C_2\), ..., \(C_n\) as follows:

- \(C_0 = \binom{n}{0} = 1\)
- \(C_1 = \binom{n}{1} = n\)
- \(C_2 = \binom{n}{2} = \frac{n(n-1)}{2}\)
- ...
- \(C_n = \binom{n}{n} = 1\)

### Step 2: Substitute the coefficients into the expression

Now we substitute these coefficients into the expression:

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

This becomes:

\[ (1 + \frac{1}{n})(1 + \frac{\frac{n(n-1)}{2}}{n})...(1 + \frac{1}{\binom{n}{n-1}}) \]

### Step 3: Simplify each term

We can simplify each term in the product:

1. The first term is \(1 + \frac{1}{n} = \frac{n+1}{n}\).
2. The second term is \(1 + \frac{(n-1)}{2} = \frac{n+1}{2}\).
3. Continuing this, we can express the general term as:

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

### Step 4: Write the product of all terms

The product of all these terms from \(r = 1\) to \(n\) is:

\[ \prod_{r=1}^{n} \left(1 + \frac{C_r}{C_{r-1}}\right) = \prod_{r=1}^{n} \frac{n+1}{r} = \frac{(n+1)^n}{n!} \]

### Conclusion

Thus, the final result of the expression is:

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