If `S_(n) = underset (r=0) overset( n) sum (1) /(""^(n) C_(r))` and `T_(n) = underset(r=0) overset(n) sum (r )/(""^(n) C_(r))` then ` (t_(n))/(s_(n))` = ?

Options

A. `1/4 n `
B. `1/3 n`
C. `1/2 n`
D. `n`

Answer

To solve the problem, we need to evaluate the expressions \( S_n \) and \( T_n \) and then find the ratio \( \frac{T_n}{S_n} \).

### Step 1: Evaluate \( S_n \)

The expression for \( S_n \) is given by:

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

Using the property of binomial coefficients, we know that:

\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]

Thus, we can rewrite \( S_n \):

\[ S_n = \sum_{r=0}^{n} \frac{r!(n-r)!}{n!} \]

### Step 2: Simplify \( S_n \)

Now, we can simplify \( S_n \):

\[ S_n = \frac{1}{n!} \sum_{r=0}^{n} r!(n-r)! \]

Notice that \( r!(n-r)! \) counts the number of ways to arrange \( n \) items, which is equal to \( n! \). Therefore, we have:

\[ S_n = \frac{1}{n!} \cdot n! = 1 \]

### Step 3: Evaluate \( T_n \)

Now, we evaluate \( T_n \):

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

Using the property of binomial coefficients again, we can rewrite \( T_n \):

\[ T_n = \sum_{r=1}^{n} \frac{r}{\frac{n!}{r!(n-r)!}} = \sum_{r=1}^{n} \frac{r! (n-r)!}{(n-1)!} \]

### Step 4: Simplify \( T_n \)

We can simplify \( T_n \):

\[ T_n = \frac{1}{(n-1)!} \sum_{r=1}^{n} r!(n-r)! \]

Notice that \( r!(n-r)! \) can be rewritten as \( (n-1)! \) when we sum over all \( r \):

\[ T_n = \frac{1}{(n-1)!} \cdot (n-1)! = n \]

### Step 5: Find \( \frac{T_n}{S_n} \)

Now we can find the ratio:

\[ \frac{T_n}{S_n} = \frac{n}{1} = n \]

### Final Answer

Thus, the final answer is:

\[ \frac{T_n}{S_n} = n \]