Question

If ${\mathrm{S}}_{\mathrm{n}}=\sum _{\mathrm{r}=0}^{\mathrm{n}}\frac{1}{{\mathrm{n}}_{\mathrm{T}}}$ and ${\mathrm{t}}_{\mathrm{n}}=\sum _{\mathrm{r}=0}^{\mathrm{n}}\frac{\mathrm{r}}{{\mathrm{n}}_{\mathrm{T}}}$ then $\frac{{\text{t}}_n}{{\mathrm{S}}_{\mathrm{n}}}$ is equal to:

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

Answer 1

The correct option is A

$\frac{\mathrm{n}}{2}$

Explanation for the correct option.

Given that, ${\mathrm{S}}_{\mathrm{n}}=\sum _{\mathrm{r}=0}^{\mathrm{n}}\frac{1}{{\mathrm{n}}_r}$ and ${\mathrm{t}}_{\mathrm{n}}=\sum _{\mathrm{r}=0}^{\mathrm{n}}\frac{\mathrm{r}}{{\mathrm{n}}_r}$.

Evaluate $t_n$:

$$\begin{gathered} \Rightarrow {\mathrm{t}}_{\mathrm{n}}=\sum _{\mathrm{r}=0}^{\mathrm{n}}\frac{\mathrm{r}}{{\mathrm{n}}_{\mathrm{r}}} \\ \Rightarrow t_n=\sum _{1=0}^n\frac{n-(n-r)}{{}^{n}C_r} \\ \Rightarrow t_n=\sum _{r=0}^n\frac{n-(n-r)}{{}^{n}C_r} \\ \Rightarrow t_n=\sum _{1=0}^n\frac{n}{{}^{n}C_r}-\frac{(n-r)}{{}^{n}C_r} \\ \Rightarrow t_n=n\sum _{r=0}^n\frac{n}{{}^{n}C_r}-\frac{\left(r\right)}{{}^{n}C_r}\because n-r=r \\ \Rightarrow t_n=n·S_n-t_n \\ \Rightarrow \frac{t_n}{s_n}=\frac{n}{2} \end{gathered}$$

Hence, option A is correct.