If `sum_(r=0)^(n) (r)/(""^(n)C_(r))= sum_(r=0)^(n) (n^(2)-3n+3)/(2.""^(n)C_(r))`, then

Options

A. a. `n = 1`
B. b. `n = 2`
C. c. `n = 3`
D. d. none of these

Explanation

To solve the equation

\[ \sum_{r=0}^{n} \frac{r}{\binom{n}{r}} = \sum_{r=0}^{n} \frac{n^2 - 3n + 3}{2 \binom{n}{r}}, \]

we will follow these steps:

### Step 1: Analyze the Left-Hand Side (LHS)

The left-hand side is

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

Using the identity \(\binom{n}{r} = \binom{n}{n-r}\), we can rewrite the sum:

\[ \sum_{r=0}^{n} \frac{r}{\binom{n}{r}} = \sum_{r=0}^{n} \frac{n - (n - r)}{\binom{n}{r}} = \sum_{r=0}^{n} \frac{n}{\binom{n}{r}} - \sum_{r=0}^{n} \frac{n - r}{\binom{n}{r}}. \]

### Step 2: Simplify the LHS

We can denote the LHS as \(k\):

\[ k = \sum_{r=0}^{n} \frac{n}{\binom{n}{r}} - k. \]

This implies:

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

Thus,

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

### Step 3: Analyze the Right-Hand Side (RHS)

The right-hand side is

\[ \sum_{r=0}^{n} \frac{n^2 - 3n + 3}{2 \binom{n}{r}}. \]

We can factor out the constant term:

\[ \frac{n^2 - 3n + 3}{2} \sum_{r=0}^{n} \frac{1}{\binom{n}{r}}. \]

### Step 4: Equate LHS and RHS

Setting the two sides equal gives:

\[ \frac{n}{2} \sum_{r=0}^{n} \frac{1}{\binom{n}{r}} = \frac{n^2 - 3n + 3}{2} \sum_{r=0}^{n} \frac{1}{\binom{n}{r}}. \]

Assuming \(\sum_{r=0}^{n} \frac{1}{\binom{n}{r}} \neq 0\), we can cancel this term:

\[ n = n^2 - 3n + 3. \]

### Step 5: Rearranging the Equation

Rearranging gives:

\[ 0 = n^2 - 4n + 3. \]

### Step 6: Factor the Quadratic

Factoring the quadratic equation:

\[ 0 = (n - 1)(n - 3). \]

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

Setting each factor to zero gives us:

\[ n - 1 = 0 \quad \Rightarrow \quad n = 1, \]
\[ n - 3 = 0 \quad \Rightarrow \quad n = 3. \]

### Conclusion

The solutions for \(n\) are \(n = 1\) and \(n = 3\).