`sum_(r = 0)^(n-1) (C_r)/(C_r + C_(r+1)) = `

Options

A. `n/2`
B. `n/3`
C. `n/4`
D. `(2n)/(3)`

Answer

To solve the problem \( \sum_{r=0}^{n-1} \frac{C_r}{C_r + C_{r+1}} \), where \( C_r \) represents the binomial coefficient \( \binom{n-1}{r} \), we can follow these steps:

### Step 1: Rewrite the expression
We start with the given summation:
\[ \sum_{r=0}^{n-1} \frac{C_r}{C_r + C_{r+1}} = \sum_{r=0}^{n-1} \frac{\binom{n-1}{r}}{\binom{n-1}{r} + \binom{n-1}{r+1}} \]

### Step 2: Simplify the denominator
Using the property of binomial coefficients, we know that:
\[ \binom{n-1}{r} + \binom{n-1}{r+1} = \binom{n}{r+1} \]
Thus, we can rewrite the expression as:
\[ \sum_{r=0}^{n-1} \frac{\binom{n-1}{r}}{\binom{n}{r+1}} \]

### Step 3: Rewrite the fraction
Now, we can express the fraction:
\[ \frac{\binom{n-1}{r}}{\binom{n}{r+1}} = \frac{\binom{n-1}{r}}{\frac{n}{r+1} \binom{n-1}{r}} = \frac{r+1}{n} \]
This means we can rewrite our summation as:
\[ \sum_{r=0}^{n-1} \frac{r+1}{n} \]

### Step 4: Factor out the constant
Since \( \frac{1}{n} \) is a constant, we can factor it out of the summation:
\[ \frac{1}{n} \sum_{r=0}^{n-1} (r+1) \]

### Step 5: Evaluate the summation
The summation \( \sum_{r=0}^{n-1} (r+1) \) can be simplified:
\[ \sum_{r=0}^{n-1} (r+1) = 1 + 2 + 3 + \ldots + n = \frac{n(n+1)}{2} \]

### Step 6: Combine the results
Substituting this back into our expression, we have:
\[ \frac{1}{n} \cdot \frac{n(n+1)}{2} = \frac{n+1}{2} \]

### Final Result
Thus, the final result of the summation is:
\[ \sum_{r=0}^{n-1} \frac{C_r}{C_r + C_{r+1}} = \frac{n}{2} \]