<p>If \(S_n = \displaystyle\sum_{r=0}^{n} \dfrac{1}{{}^nC_r}\) and \(t_n = \displaystyle\sum_{r=0}^{n} \dfrac{r}{{}^nC_r}\), then \(\dfrac{t_n}{S_n}\) is equal to</p>
Step-by-Step Solution
Key Concept: Use the symmetry property of binomial coefficients: C(n,r) = C(n,n-r). By pairing terms in the sum for t_n with their symmetric counterparts and leveraging the symmetry of the numerator r, the ratio simplifies to a constant independent of specific binomial coefficients.
<p><strong>Step 1:</strong> Use the symmetry property C(n,r) = C(n,n-r) to pair terms in t_n.</p><p><strong>Step 2:</strong> For each pair of terms with indices r and (n-r), we have:</p><p>$$\frac{r}{\binom{n}{r}} + \frac{n-r}{\binom{n}{n-r}} = \frac{r}{\binom{n}{r}} + \frac{n-r}{\binom{n}{r}} = \frac{n}{\binom{n}{r}}$$</p><p><strong>Step 3:</strong> Summing all such pairs:</p><p>$$t_n = \sum_{r=0}^{n} \frac{r}{\binom{n}{r}} = \sum_{r=0}^{\lfloor n/2 \rfloor} \left(\frac{n}{\binom{n}{r}}\right) = \frac{n}{2} \sum_{r=0}^{n} \frac{1}{\binom{n}{r}} = \frac{n}{2} S_n$$</p><p><strong>Step 4:</strong> Therefore:</p><p>$$\frac{t_n}{S_n} = \frac{\frac{n}{2}S_n}{S_n} = \frac{n}{2}$$</p><p>∴ Answer: A</p>
Correct Answer: A