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

To solve the given problem, we need to evaluate the expression:

\[\cos^{-1}\left( \frac{S_n}{n t_n} \right)\]

where

\(t_n = \sum_{r=0}^n \frac{1}{\left({}^nC_r\right)^k}\)

and

\(S_n = \sum_{r=0}^n \frac{r}{\left({}^nC_r\right)^k}\)

These are summations over binomial coefficients raised to the power of \(k\). Let us analyze the given functions:

Step-by-Step Calculation and Analysis

1. First, note that the binomial coefficient \({}^nC_r\) can be expressed as \(\frac{n!}{r!(n-r)!}\).
2. In \(t_n\), each term is \(\frac{1}{\left({}^nC_r\right)^k} = \frac{(r!(n-r)!)^k}{(n!)^k}\).
3. For \(S_n\), each term has an additional factor of \(r\) in the numerator, i.e., \(\frac{r}{\left({}^nC_r\right)^k} = \frac{r(r!(n-r)!)^k}{(n!)^k}\).
4. Now, consider the expression \(\frac{S_n}{n t_n}\):
5. The term \(\frac{S_n}{t_n}\) simplifies to a weighted average of \(r\) values in sequential order over \(n\) terms, resembling a central binomial distribution pattern.
6. Without the constraints directly on \(\cos^{-1}\) values, based on mathematical properties, we infer that the numerator and denominator balance around a significant symmetry. Given the uniformity in \(n\) and an averaging mechanism between \(0\) and \(n\), it indicates a resultant structure of \(0\) centered on symmetry in averages.
7. The angle that correlates with a cosine of zero is \(\frac{\pi}{2}\), implying the resultant value.

Conclusion

Thus, the resultant value of \(\cos^{-1}\left( \frac{S_n}{n t_n} \right)\) simplifies to:

\(\frac{\pi}{2}\)