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 [AIEEE 2004]

If \[{{S}_{n}}=\sum\limits_{r=0}^{n}{\frac{1}{^{n}{{C}_{r}}}}\] and \[

1.	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 [AIEEE 2004]
A.	\[\frac{2n-1}{2}\]
B.	\[\frac{1}{2}n-1\]
C.	\[n-1\]
D.	\[\frac{1}{2}n\]
Answer» E.

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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 [AIEEE 2004]

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