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

Answer

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

- \( s_n = \sum_{r=0}^{n} \frac{1}{\binom{n}{r}} \)
- \( t_n = \sum_{r=0}^{n} \frac{r}{\binom{n}{r}} \)

### Step-by-Step Solution:

1. **Express \( t_n \) in a different form:**
We can rewrite \( t_n \) by adding and subtracting \( n \) in the numerator:
\[ t_n = \sum_{r=0}^{n} \frac{r}{\binom{n}{r}} = \sum_{r=0}^{n} \frac{n - (n - r)}{\binom{n}{r}} = \sum_{r=0}^{n} \frac{n}{\binom{n}{r}} - \sum_{r=0}^{n} \frac{n - r}{\binom{n}{r}} \]

2. **Simplify the first term:**
The first term simplifies to:
\[ t_n = n \sum_{r=0}^{n} \frac{1}{\binom{n}{r}} - \sum_{r=0}^{n} \frac{n - r}{\binom{n}{r}} \]
We know that \( \sum_{r=0}^{n} \frac{1}{\binom{n}{r}} = s_n \).

3. **Transform the second term:**
For the second term, we can use the property of binomial coefficients:
\[ \sum_{r=0}^{n} \frac{n - r}{\binom{n}{r}} = \sum_{r=0}^{n} \frac{n}{\binom{n}{r}} - \sum_{r=0}^{n} \frac{r}{\binom{n}{r}} = n s_n - t_n \]
Thus, we can write:
\[ t_n = n s_n - (n s_n - t_n) \]

4. **Combine and solve for \( t_n \):**
Rearranging gives:
\[ t_n + t_n = n s_n \implies 2t_n = n s_n \implies t_n = \frac{n}{2} s_n \]

5. **Find the ratio \( \frac{t_n}{s_n} \):**
Now, we can find the ratio:
\[ \frac{t_n}{s_n} = \frac{\frac{n}{2} s_n}{s_n} = \frac{n}{2} \]

### Final Result:
Thus, the ratio \( \frac{t_n}{s_n} \) is equal to \( \frac{n}{2} \).