If `s_(n)=sum_(r=0)^(n)(1)/(.^(n)C_(r))and t_(n)=sum_(r=0)^(n)(r)/(.^(n)C_(r))`, then `(t_(n))/(s_(n))` is equal to

Options

A. `n-1`
B. `(1)/(2)n-1`
C. `(1)/(2)n`
D. `(2n-1)/(2)`

Answer

To solve the problem, we need to find the value of \( \frac{t_n}{s_n} \) where

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

and

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

### Step 1: Express \( t_n \) in a different form

We start with the expression for \( t_n \):

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

We can rewrite \( r \) as \( n - (n - r) \):

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

### Step 2: Use the symmetry property of binomial coefficients

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

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

This means we can express \( t_n \) as:

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

### Step 3: Combine the two expressions

Adding both expressions for \( t_n \):

\[ 2t_n = \sum_{r=0}^{n} \left( \frac{r}{\binom{n}{r}} + \frac{n - r}{\binom{n}{r}} \right) = \sum_{r=0}^{n} \frac{n}{\binom{n}{r}}. \]

Thus, we have:

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

### Step 4: Solve for \( t_n \)

From the equation \( 2t_n = n s_n \), we can express \( t_n \):

\[ t_n = \frac{n s_n}{2}. \]

### Step 5: Calculate \( \frac{t_n}{s_n} \)

Now, we can find \( \frac{t_n}{s_n} \):

\[ \frac{t_n}{s_n} = \frac{\frac{n s_n}{2}}{s_n} = \frac{n}{2}. \]

### Final Answer

Thus, the value of \( \frac{t_n}{s_n} \) is:

\[ \frac{t_n}{s_n} = \frac{n}{2}. \]