If `S_n=sum_(r=0)^n 1/(nC_r) and t_n=sum_(r=0)^n r/(nC_r),` then `t_n/S_n=`

Options

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

Explanation

To solve the problem, we need to find the ratio \( \frac{t_n}{S_n} \) 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 \( t_n \)

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

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

We can rewrite \( r \) in terms of \( n \) and \( \binom{n}{r} \):

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

### Step 2: Change the Index of Summation

Next, we change the index of summation. Let \( k = r - 1 \), then when \( r = 0 \), \( k = -1 \) (which we ignore) and when \( r = n \), \( k = n - 1 \):

\[ t_n = n \sum_{k=0}^{n-1} \frac{1}{\binom{n-1}{k}}. \]

### Step 3: Relate to \( S_n \)

Now, we can express \( S_n \):

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

Notice that:

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

### Step 4: Combine the Results

Now we have:

\[ t_n = n \sum_{k=0}^{n-1} \frac{1}{\binom{n-1}{k}}. \]

Thus, we can express \( S_n \) as:

\[ S_n = \sum_{k=0}^{n-1} \frac{1}{\binom{n-1}{k}} + 1. \]

### Step 5: Find the Ratio

Now we can find the ratio \( \frac{t_n}{S_n} \):

\[ \frac{t_n}{S_n} = \frac{n \sum_{k=0}^{n-1} \frac{1}{\binom{n-1}{k}}}{\sum_{k=0}^{n-1} \frac{1}{\binom{n-1}{k}} + 1}. \]

### Step 6: Simplify the Expression

Let \( x = \sum_{k=0}^{n-1} \frac{1}{\binom{n-1}{k}} \):

\[ \frac{t_n}{S_n} = \frac{n x}{x + 1}. \]

### Step 7: Evaluate the Limit

As \( n \) approaches infinity, we can analyze the behavior of \( x \) and conclude that:

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

Thus, the final answer is:

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

### Final Answer

The final result is:

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