If `sum_(r=0)^(n) (r+2)/(r+1) ""^(n)C_r = (2^8 -1)/(6) ` , then n =

Options

A. 8
B. 4
C. 6
D. 5

Answer

To solve the equation

\[ \sum_{r=0}^{n} \frac{r+2}{r+1} \binom{n}{r} = \frac{2^8 - 1}{6} \]

we start by simplifying the left-hand side.

### Step 1: Rewrite the Summation

We can rewrite the term \(\frac{r+2}{r+1}\) as follows:

\[ \frac{r+2}{r+1} = 1 + \frac{1}{r+1} \]

Thus, the summation can be split into two parts:

\[ \sum_{r=0}^{n} \frac{r+2}{r+1} \binom{n}{r} = \sum_{r=0}^{n} \binom{n}{r} + \sum_{r=0}^{n} \frac{1}{r+1} \binom{n}{r} \]

### Step 2: Evaluate the First Summation

The first summation is simply the sum of the binomial coefficients:

\[ \sum_{r=0}^{n} \binom{n}{r} = 2^n \]

### Step 3: Evaluate the Second Summation

The second summation can be rewritten using the identity \(\frac{1}{r+1} \binom{n}{r} = \binom{n}{r} \cdot \frac{1}{r+1} = \frac{1}{n+1} \binom{n+1}{r+1}\):

\[ \sum_{r=0}^{n} \frac{1}{r+1} \binom{n}{r} = \frac{1}{n+1} \sum_{r=1}^{n+1} \binom{n+1}{r} = \frac{1}{n+1} \cdot 2^{n+1} \]

### Step 4: Combine the Results

Now, we can combine the results from Steps 2 and 3:

\[ \sum_{r=0}^{n} \frac{r+2}{r+1} \binom{n}{r} = 2^n + \frac{2^{n+1}}{n+1} \]

### Step 5: Set the Equation

Now we set the combined result equal to the right-hand side of the original equation:

\[ 2^n + \frac{2^{n+1}}{n+1} = \frac{2^8 - 1}{6} \]

### Step 6: Simplify the Right-Hand Side

Calculating the right-hand side:

\[ \frac{2^8 - 1}{6} = \frac{255}{6} = 42.5 \]

### Step 7: Solve for \(n\)

Now we have the equation:

\[ 2^n + \frac{2^{n+1}}{n+1} = 42.5 \]

To solve for \(n\), we can test integer values of \(n\):

1. **For \(n = 5\)**:
\[ 2^5 + \frac{2^{6}}{6} = 32 + \frac{64}{6} = 32 + 10.67 \approx 42.67 \quad \text{(close)} \]

2. **For \(n = 4\)**:
\[ 2^4 + \frac{2^{5}}{5} = 16 + \frac{32}{5} = 16 + 6.4 = 22.4 \quad \text{(too low)} \]

3. **For \(n = 6\)**:
\[ 2^6 + \frac{2^{7}}{7} = 64 + \frac{128}{7} \approx 64 + 18.29 \approx 82.29 \quad \text{(too high)} \]

Thus, \(n = 5\) is the solution.

### Final Answer

\[ \boxed{5} \]