If {S_n} = sumlimits_{r = 0}^n {frac{1}{{^n{C | vedclass

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(D) We have ${S_n} = \sum\limits_{r = 0}^n {\frac{1}{{^n{C_r}}}} $ and ${t_n} = \sum\limits_{r = 0}^n {\frac{r}{{^n{C_r}}}} $. Using the property ${^n

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$\frac{1}{2}n$

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(D) We have ${S_n} = \sum\limits_{r = 0}^n {\frac{1}{{^n{C_r}}}} $ and ${t_n} = \sum\limits_{r = 0}^n {\frac{r}{{^n{C_r}}}} $. Using the property ${^n{C_r}} = {^n{C_{n - r}}}$,we can write ${t_n} = \sum\limits_{r = 0}^n {\frac{{n - r}}{{^n{C_{n - r}}}}} = \sum\limits_{r = 0}^n {\frac{{n - r}}{{^n{C_r}}}} $. Thus,${t_n} = n \sum\limits_{r = 0}^n {\frac{1}{{^n{C_r}}}} - \sum\limits_{r = 0}^n {\frac{r}{{^n{C_r}}}} $. This simplifies to ${t_n} = n \cdot {S_n} - {t_n}$. Adding ${t_n}$ to both sides,we get $2{t_n} = n \cdot {S_n}$. Therefore,$\frac{{{t_n}}}{{{S_n}}} = \frac{n}{2}$.

If ${S_n} = \sum\limits_{r = 0}^n {\frac{1}{{^n{C_r}}}} $ and ${t_n} = \sum\limits_{r = 0}^n {\frac{r}{{^n{C_r}}}} $,then $\frac{{{t_n}}}{{{S_n}}}$ is equal to

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