If `s_n=sum_(r < s) (1/(nC_r)+1/(nC_s)) and t_n=sum_(r < s)(r/(nC_r)+s/(nC_s)),` then `t_n/s_n=`

Options

A. n-1
B. n+1
C. `(n)/(2)`
D. none of these

Explanation

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

\[ s_n = \sum_{r=0}^{s-1} \left( \frac{1}{\binom{n}{r}} + \frac{1}{\binom{n}{s}} \right) \]
\[ t_n = \sum_{r=0}^{s-1} \left( \frac{r}{\binom{n}{r}} + \frac{s}{\binom{n}{s}} \right) \]

### Step 1: Compute \( s_n \)

1. **Substitute \( s = n \)** in \( s_n \):
\[ s_n = \sum_{r=0}^{n-1} \left( \frac{1}{\binom{n}{r}} + \frac{1}{\binom{n}{n}} \right) \]
Since \( \binom{n}{n} = 1 \), we have:
\[ s_n = \sum_{r=0}^{n-1} \frac{1}{\binom{n}{r}} + 1 \]

2. **Recognize that \( \sum_{r=0}^{n-1} \frac{1}{\binom{n}{r}} \)** can be simplified:
\[ s_n = \sum_{r=0}^{n-1} \frac{1}{\binom{n}{r}} + 1 = n \cdot \frac{1}{\binom{n}{n}} = n \]
Thus, we can write:
\[ s_n = n + 1 \]

### Step 2: Compute \( t_n \)

1. **Substitute \( s = n \)** in \( t_n \):
\[ t_n = \sum_{r=0}^{n-1} \left( \frac{r}{\binom{n}{r}} + \frac{n}{\binom{n}{n}} \right) \]
Again, since \( \binom{n}{n} = 1 \):
\[ t_n = \sum_{r=0}^{n-1} \frac{r}{\binom{n}{r}} + n \]

2. **Recognize that \( \sum_{r=0}^{n-1} \frac{r}{\binom{n}{r}} \)** can also be simplified:
\[ t_n = n \cdot \sum_{r=0}^{n-1} \frac{1}{\binom{n-1}{r-1}} + n \]
This is equivalent to:
\[ t_n = n \cdot (s_n - 1) + n \]
Thus, we can write:
\[ t_n = n \cdot n + n = n^2 + n \]

### Step 3: Compute the ratio \( \frac{t_n}{s_n} \)

Now we can compute the ratio:
\[ \frac{t_n}{s_n} = \frac{n^2 + n}{n + 1} \]

### Step 4: Simplify the ratio

1. **Factor out \( n \)** from the numerator:
\[ \frac{n(n + 1)}{n + 1} \]
Cancel \( n + 1 \) from the numerator and denominator:
\[ \frac{t_n}{s_n} = n \]

### Final Result

Thus, the final answer is:
\[ \frac{t_n}{s_n} = n \]