Question

If \[ t_n = \sum_{r=0}^n \frac{1}{\left({}^nC_r\right)^k} \quad \text{and} \quad S_n = \sum_{r=0}^n \frac{r}{\left({}^nC_r\right)^k}, \] where \( k \in \mathbb{Z}^+ \), then \[ \cos^{-1}\left( \frac{S_n}{n t_n} \right) \] is:

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

Hint:

The symmetry property of binomial coefficients \(\binom{n}{r} = \binom{n}{n-r}\) is key to solving such sums.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Use the property that \({}^nC_r = {}^nC_{n-r}\) to relate \(S_n\) and \(t_n\).

Step 2: Detailed Explanation:
Write \(S_n = \sum_{r=0}^n \frac{r}{({}^nC_r)^k}\). Replace \(r\) by \(n-r\): \[ S_n = \sum_{r=0}^n \frac{n-r}{({}^nC_{n-r})^k} = \sum_{r=0}^n \frac{n-r}{({}^nC_r)^k} \] Adding the two expressions for \(S_n\): \[ 2S_n = \sum_{r=0}^n \frac{r + n - r}{({}^nC_r)^k} = n \sum_{r=0}^n \frac{1}{({}^nC_r)^k} = n t_n \] Thus \(\frac{S_n}{n t_n} = \frac{1}{2}\). Then \(\cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3}\). Wait, there is a mis-step: The sum \(S_n\) is defined with \(r\) in numerator. The property \({}^nC_r = {}^nC_{n-r}\) gives: \[ S_n = \sum_{r=0}^n \frac{r}{({}^nC_r)^k} = \sum_{r=0}^n \frac{n-r}{({}^nC_{n-r})^k} = \sum_{r=0}^n \frac{n-r}{({}^nC_r)^k} \] Adding: \(2S_n = \sum_{r=0}^n \frac{n}{({}^nC_r)^k} = n t_n\). So \(\frac{S_n}{n t_n} = \frac{1}{2}\). Thus \(\cos^{-1}(1/2) = \pi/3\). But the answer given in many sources is \(\pi/2\). Let's re-check: If \(S_n = \sum r/(C_r)^k\) and \(t_n = \sum 1/(C_r)^k\), then \(S_n/t_n\) is the average of \(r\), which for symmetric binomial coefficients is \(n/2\). So \(S_n/(n t_n) = 1/2\). Then \(\cos^{-1}(1/2) = \pi/3\). However, the problem likely intends \(\cos^{-1}(S_n/(n t_n)) = \pi/3\). But the given answer choices have \(\pi/2\) as well. Re-evaluate: There is a known identity: \(\sum r/{}^nC_r = n/2 \sum 1/{}^nC_r\). So ratio is \(n/2\). Then \(\cos^{-1}((n/2)/(n)) = \cos^{-1}(1/2) = \pi/3\). But if the denominator is \(n t_n\), then it's \(\cos^{-1}((n t_n/2)/(n t_n)) = \cos^{-1}(1/2) = \pi/3\). So answer should be \(\pi/3\). But the provided answer key says \(\pi/2\). Possibly a misprint. Following the derivation: \(S_n = \sum r a_r\), \(t_n = \sum a_r\) with \(a_r = 1/({}^nC_r)^k\). Since \(a_r = a_{n-r}\), \(S_n = \sum (n-r)a_r = n t_n - S_n \Rightarrow 2S_n = n t_n \Rightarrow S_n/(n t_n) = 1/2\). So \(\cos^{-1}(1/2) = \pi/3\). I will present the derived answer.

Step 3: Final Answer:
\(\frac{S_n}{n t_n} = \frac{1}{2}\), so \(\cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3}\), which corresponds to option (B).