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

Answer 1

Answer: (A)

$\frac{n}{2}$

$S_n={\sum }_{r=0}^n\frac{1}{{}^n{C}_r}$ and $t_n={\sum }_{r=0}^n\frac{r}{{}^n{C}_r}$

$t_n={\sum }_{r=0}^n\frac{r}{{}^n{C}_r}$

$t_n={\sum }_{r=0}^n\frac{n-(n-r)}{{}^n{C}_r}$

$t_n={\sum }_{r=0}^n\frac{n-(n-r)}{{}^n{C}_r}$

$t_n={\sum }_{r=0}^n\frac{n}{{}^n{C}_r}-\frac{(n-r)}{{}^n{C}_r}$

$t_n=n{\sum }_{r=0}^n\frac{n}{{}^1{C}_r}-\frac{\left(r\right)}{{}^n{C}_r}\because n-r=r$

$t_n=n\cdot S_n-t_n$

$\frac{t_n}{S_n}=\frac{n}{2}$