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If s(n)=sum(r=0)^(n)(1)/(.^(n)C(r))and t...

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

A

`n-1`

B

`(1)/(2)n-1`

C

`(1)/(2)n`

D

`(2n-1)/(2)`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to evaluate the expressions for \( s_n \) and \( t_n \) and then find the ratio \( \frac{t_n}{s_n} \). ### Step 1: Define \( s_n \) The expression for \( s_n \) is given by: \[ s_n = \sum_{r=0}^{n} \frac{1}{\binom{n}{r}} \] This represents the sum of the reciprocals of the binomial coefficients. ### Step 2: Define \( t_n \) The expression for \( t_n \) is given by: \[ t_n = \sum_{r=0}^{n} \frac{r}{\binom{n}{r}} \] This represents the sum of the terms where each term is the index \( r \) multiplied by the reciprocal of the binomial coefficient. ### Step 3: Simplify \( t_n \) We can rewrite \( t_n \) as follows: \[ t_n = \sum_{r=0}^{n} \frac{r}{\binom{n}{r}} = \sum_{r=1}^{n} \frac{r}{\binom{n}{r}} = \sum_{r=1}^{n} \frac{n}{n} \cdot \frac{r}{\binom{n}{r}} = n \sum_{r=1}^{n} \frac{1}{\binom{n-1}{r-1}} \] This is because \( r \cdot \binom{n}{r} = n \cdot \binom{n-1}{r-1} \). ### Step 4: Relate \( t_n \) to \( s_n \) Now, we can express \( t_n \) in terms of \( s_n \): \[ t_n = n \sum_{r=0}^{n-1} \frac{1}{\binom{n-1}{r}} = n s_{n-1} \] where \( s_{n-1} = \sum_{r=0}^{n-1} \frac{1}{\binom{n-1}{r}} \). ### Step 5: Find the ratio \( \frac{t_n}{s_n} \) Now we can find the ratio: \[ \frac{t_n}{s_n} = \frac{n s_{n-1}}{s_n} \] ### Step 6: Evaluate \( s_n \) From the properties of binomial coefficients, we know that: \[ s_n = \sum_{r=0}^{n} \frac{1}{\binom{n}{r}} = \frac{2^n}{\binom{n}{0}} = 2^n \] Thus, we can write: \[ s_n = 2^n \] ### Step 7: Substitute \( s_n \) into the ratio Now substituting \( s_n \) into the ratio gives: \[ \frac{t_n}{s_n} = \frac{n s_{n-1}}{2^n} \] ### Step 8: Evaluate \( s_{n-1} \) Similarly, we can evaluate \( s_{n-1} \): \[ s_{n-1} = 2^{n-1} \] So, substituting this into our ratio: \[ \frac{t_n}{s_n} = \frac{n \cdot 2^{n-1}}{2^n} = \frac{n}{2} \] ### Final Answer Thus, we conclude that: \[ \frac{t_n}{s_n} = \frac{n}{2} \]
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