Question

If $t_n={\sum }_{r=0}^n\frac{1}{{\left({{}^{n}C}_r\right)}^k}$ and $S_n={\sum }_{r=0}^n\frac{r}{{\left({{}^{n}C}_r\right)}^k}$, where $k\in Z^{+}$, then ${\cos }^{-1}\left(\frac{S_n}{n{t}_n}\right)$

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{2}$

Answer 1

The correct option is B $\frac{\pi }{4}$

$S_n=\frac{0}{{\left({{}^{n}C}_0\right)}^k}+\frac{1}{{\left({{}^{n}C}_1\right)}^k}+\frac{2}{{\left({{}^{n}C}_2\right)}^k}+...\frac{n}{{\left({{}^{n}C}_n\right)}^k}$
$\Rightarrow S_n=\frac{n}{{\left({{}^{n}C}_0\right)}^k}+\frac{n-1}{{\left({{}^{n}C}_0\right)}^k}+\frac{n-2}{{\left({{}^{n}C}_0\right)}^k}+....+\frac{0}{{\left({{}^{n}C}_0\right)}^k}$
$\Rightarrow {2S}_n=n{t}_n$
$\Rightarrow \frac{S_n}{n{t}_n}=\frac{1}{2}$
Now, ${\cos }^{-1}\left(\frac{S_n}{n{t}_n}\right)={\cos }^{-1}\left(\frac{1}{2}\right)=\frac{\pi }{3}$