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If $s_{n}=\sum_{r=0}^{n}\frac{1}{^{n}C_{r}}\; and\; \, t_{n}=\sum_{r=0}^{n}\frac{r}{^{n}C_{r}}\; ,$ then $\frac{t_{n}}{s_{n}}$ is equal to

Option 1)
$n-1\; \;$

Option 2)
$\; \frac{1}{2}n-1\;$

Option 3)
$\; \frac{1}{2}n\;$

Option 4)
$\; \frac{2n-1}{2}$

Answers (1)

best_answer

As we have learned

Properties of Binomial Theorem -

Binomial coefficients equidistant from beginning and end are equal.

- wherein

$\dpi{120} ^{n}c_{r}= ^{n}c_{n-r}$

$S_n = \frac{1}{^nC_0}+ \frac{1}{^nC_1}+ \frac{1}{^nC_2}+.....+ \frac{1}{^nC_n}$

$t_n = \frac{1}{^nC_1}+ \frac{1}{^nC_2}+.....+ \frac{n}{^nC_n}$

$= \frac{1}{^nC_{n-1}}+ \frac{1}{^nC_{n-2}}+.....+ \frac{n}{^nC_0}$ (Using $^nC_r = ^nC_{n-r}$ )

We have

$= \frac{n}{^nC_0}+ \frac{n}{^nC_1}+ \frac{n}{^nC_2}+.....+ \frac{n}{^nC_n}$

$= \frac{1}{^nC_0}+ (\frac{n-1}{^nC_1}+ \frac{1}{^nC_1})+( \frac{n-2}{^nC_2} + \frac{2}{^nC_2})+ ....+ (\frac{n}{^n C_n })$

$= (\frac{1}{^nC_0}+ \frac{1}{^nC_1}+..........+ \frac{n}{^nC_n})$

$= (\frac{n-1}{^nC_1}+ \frac{n-2}{^nC_2}+..........+ \frac{1}{^nC_{n-1}})$

$= (\frac{1}{^nC_0}+ t_n ) (t_n) - \frac{1}{^nC_0}= 2 t_n$

Option 1)

$n-1\; \;$

Option 2)

$\; \frac{1}{2}n-1\;$

Option 3)

$\; \frac{1}{2}n\;$

Option 4)

$\; \frac{2n-1}{2}$

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