Q57 / 170 · Manipal (MET) 2016 · Mathematics

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 $\cos ^{-1}\left(\frac{S_n}{n t_n}\right)$ is

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

Solution

$S_n=\frac{0}{\left({ }^n C_0\right)^k}+\frac{1}{\left({ }^n C_1\right)^k}+\frac{2}{\left({ }^n C_2\right)^k}+\ldots+\frac{n}{\left({ }^n C_n\right)^k}$ $\Rightarrow \quad S_n=\frac{n}{\left({ }^n C_0\right)^k}+\frac{n-1}{\left({ }^n C_1\right)^k}+\frac{n-2}{\left({ }^n C_2\right)^k}+\ldots+\frac{0}{\left({ }^n C_0\right)^k}$ $\begin{array}{ll}\Rightarrow & 2 S_n=n t_n \\ \Rightarrow & \frac{S_n}{n t_n}=\frac{1}{2}\end{array}$ Now, $\cos ^{-1}\left(\frac{S_n}{n t_n}\right)=\cos ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3}$

Hence, the correct answer is option: 'B'.

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