If $\mathrm{S}_{\mathrm{n}}=\sum_{\mathrm{r}=0}^{\mathrm{n}} \frac{1}{{}^{\mathrm{n}}\mathrm{C}_{\mathrm{r}}}$ and $\mathrm{t}_{\mathrm{n}}=\sum_{\mathrm{r}=0}^{\mathrm{n}} \frac{\mathrm{r}}{{}^{\mathrm{n}}\mathrm{C}_{\mathrm{r}}}$, then $\frac{\mathrm{t}_{\mathrm{n}}}{\mathrm{S}_{\mathrm{n}}}$ is equal to

JEE Main · 2004 · 24 Apr 2004 · Offline · Binomial Theorem

S_{n} = \sum_{r = 0}^{n} \frac{1}{^{n} C _{r}}

and

t_{n} = \sum_{r = 0}^{n} \frac{r}{^{n} C _{r}}

, then

\frac{t _{n}}{S _{n}}

is equal to

Answer: (A)

Pair each term with its mirror image using

(r n) = (n - r n)

. Replacing

r

n - r

shows

t_{n} = \sum_{r = 0}^{n} \frac{n - r}{( r n )}

as well, so adding the two forms gives

2 t_{n} = \sum_{r = 0}^{n} \frac{n}{( r n )} = n S_{n}

. Hence

\frac{t _{n}}{S _{n}} = \frac{n}{2}

JEE Main · 2004 · 24 Apr 2004 · Offline · Single correct option