Question

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 -

• A. $\frac{n}{2}$
• B. $\frac{n}{2}-1$
• C. $n-1$
• D. $\frac{2n-1}{2}$

Answer 1

The correct option is A $\frac{n}{2}$

$S_n={\sum }_{r=0}^n\frac{1}{{}^n{C}_r}=\frac{1}{{}^n{C}_0}+\frac{1}{{}^n{C}_1}+\frac{1}{{}^n{C}_2}+...+\frac{1}{{}^n{C}_{n-1}}+\frac{1}{{}^n{C}_n}{S}_n=1+\{\frac{1}{{}^n{C}_1}+\frac{1}{{}^n{C}_2}+...+\frac{1}{{}^n{C}_{n-1}}\}+1{}^n{C}_0=1,{}^n{C}_n=1,{}^n{C}_r={}^n{C}_{n-r}{S}_n=2+2\{\frac{1}{{}^n{C}_1}+\frac{1}{{}^n{C}_2}+...+\frac{1}{{}^n{C}_{n/2}}\}\frac{S_n-2}{2}=\{\frac{1}{{}^n{C}_1}+\frac{1}{{}^n{C}_2}+...+\frac{1}{{}^n{C}_{n/2}}\}\left(1\right)t_n={\sum }_{r=0}^n\frac{r}{{}^n{C}_r}=0+\frac{1}{{}^n{C}_1}+\frac{2}{{}^n{C}_2}+\frac{3}{{}^n{C}_3}+...+\frac{n-2}{{}^n{C}_{n-2}}+\frac{n-1}{{}^n{C}_{n-1}}+\frac{n}{{}^n{C}_n}{t}_n=\frac{1+n-1}{{}^n{C}_1}+\frac{2+n-2}{{}^n{C}_2}+...+\frac{n/2+n-n/2}{{}^n{C}_{n/2}}+\frac{n}{1}{t}_n=n[\frac{1}{{}^n{C}_1}+\frac{1}{{}^n{C}_2}+...+\frac{1}{{}^n{C}_{n/2}}]+n$

From $\left(1\right)$

$t_n=n\left(\frac{S_n-2}{2}\right)+n{t}_n=n\left(\frac{S_n-2+2}{2}\right)=\frac{n}{2}{S}_n\frac{t_n}{S_n}=\frac{n}{2}$