1. Key Concept: Symmetry of Binomial Coefficients

The cornerstone for solving this problem lies in a fundamental property of binomial coefficients, known as the symmetry property: $^n C_r = ^n C_{n-r}$ This property states that the number of ways to choose $r$ items from a set of $n$ is the same as the number of ways to choose $(n-r)$ items from the same set. Essentially, the sequence of binomial coefficients for a given $n$ is symmetric around its center. For example, $^n C_0 = ^n C_n$, $^n C_1 = ^n C_{n-1}$, and so on. This symmetry is extremely powerful for simplifying summations involving binomial coefficients.

2. Analyzing the Given Series

We are provided with two summation series:

• Series $S_n$: $S_n = \sum_{r=0}^n \frac{1}{^n C_r}$ This series represents the sum of the reciprocals of all binomial coefficients for a given $n$. Expanding it helps visualize the terms: $S_n = \frac{1}{^n C_0} + \frac{1}{^n C_1} + \frac{1}{^n C_2} + \dots + \frac{1}{^n C_{n-1}} + \frac{1}{^n C_n}$

• Series $t_n$: $t_n = \sum_{r=0}^n \frac{r}{^n C_r}$ This series involves $r$ in the numerator, which changes from term to term. Expanding this series will show the pattern of the numerators: $t_n = \frac{0}{^n C_0} + \frac{1}{^n C_1} + \frac{2}{^n C_2} + \dots + \frac{(n-1)}{^n C_{n-1}} + \frac{n}{^n C_n} \quad \text{(Equation 1)}$ Our ultimate goal is to find the ratio $\frac{t_n}{S_n}$.

3. The Strategy: Transforming $t_n$ using Symmetry

To simplify $t_n$ and relate it to $S_n$, we will employ the symmetry property. The key idea is to rewrite $t_n$ in a different but equivalent form.

• Step 1: Change the index of summation. We can replace the index $r$ with $(n-r)$ in the summation. This effectively reverses the order of summation, but the sum itself remains unchanged. $t_n = \sum_{r=0}^n \frac{r}{^n C_r}$ Let's change the summation variable. If we replace $r$ with $k$, then $k$ runs from $0$ to $n$. Now, let's substitute $k = n-j$. As $k$ goes from $0$ to $n$, $j$ goes from $n$ to $0$. So, $j$ also runs from $0$ to $n$. $t_n = \sum_{j=0}^n \frac{n-j}{^n C_{n-j}}$ (We can switch the dummy variable $j$ back to $r$ for consistency). $t_n = \sum_{r=0}^n \frac{n-r}{^n C_{n-r}}$ Why this step? By introducing $(n-r)$ in the numerator, we are setting up an opportunity to use the symmetry property of binomial coefficients in the denominator.

• Step 2: Apply the symmetry property. Now, we apply the property $^n C_{n-r} = ^n C_r$ to the denominator of our transformed $t_n$: $t_n = \sum_{r=0}^n \frac{n-r}{^n C_r}$ Expanding this new form of $t_n$: $t_n = \frac{n-0}{^n C_0} + \frac{n-1}{^n C_1} + \frac{n-2}{^n C_2} + \dots + \frac{n-(n-1)}{^n C_{n-1}} + \frac{n-n}{^n C_n}$ $t_n = \frac{n}{^n C_0} + \frac{n-1}{^n C_1} + \frac{n-2}{^n C_2} + \dots + \frac{1}{^n C_{n-1}} + \frac{0}{^n C_n} \quad \text{(Equation 2)}$ Why this step? Notice that the numerators in Equation 2 (i.e., $n, n-1, n-2, \dots, 1, 0$) are complementary to the numerators in Equation 1 (i.e., $0, 1, 2, \dots, n-1, n$). This complementary relationship is crucial for the next step.

4. Combining Expressions for $t_n$

Now, we add Equation 1 and Equation 2. This is a common and powerful technique when dealing with sums where the terms have a symmetric or complementary pattern.

(1) $t_n = \frac{0}{^n C_0} + \frac{1}{^n C_1} + \frac{2}{^n C_2} + \dots + \frac{(n-1)}{^n C_{n-1}} + \frac{n}{^n C_n}$ (2) $t_n = \frac{n}{^n C_0} + \frac{n-1}{^n C_1} + \frac{n-2}{^n C_2} + \dots + \frac{1}{^n C_{n-1}} + \frac{0}{^n C_n}$

Adding the two equations term by term: $2t_n = \left(\frac{0}{^n C_0} + \frac{n}{^n C_0}\right) + \left(\frac{1}{^n C_1} + \frac{n-1}{^n C_1}\right) + \dots + \left(\frac{n}{^n C_n} + \frac{0}{^n C_n}\right)$ $2t_n = \frac{0+n}{^n C_0} + \frac{1+(n-1)}{^n C_1} + \frac{2+(n-2)}{^n C_2} + \dots + \frac{(n-1)+1}{^n C_{n-1}} + \frac{n+0}{^n C_n}$ Why add them? By adding the corresponding terms, the variable parts of the numerators ($r$ and $n-r$) sum up to a constant value, $n$. This simplifies the entire summation significantly.

Simplifying each numerator, we observe that every term in the sum for $2t_n$ now has $n$ in the numerator: $2t_n = \frac{n}{^n C_0} + \frac{n}{^n C_1} + \frac{n}{^n C_2} + \dots + \frac{n}{^n C_{n-1}} + \frac{n}{^n C_n}$

5. Calculating the Ratio $\frac{t_n}{S_n}$

Now that we have a simplified expression for $2t_n$, we can relate it to $S_n$.

• Step 1: Factor out $n$. Factor out the common term $n$ from the right-hand side of the equation for $2t_n$: $2t_n = n \left( \frac{1}{^n C_0} + \frac{1}{^n C_1} + \frac{1}{^n C_2} + \dots + \frac{1}{^n C_{n-1}} + \frac{1}{^n C_n} \right)$ Why factor out $n$? This step reveals the structure of $S_n$ within the expression for $2t_n$.

• Step 2: Recognize $S_n$. Observe that the expression inside the parentheses is precisely the definition of $S_n$: $S_n = \frac{1}{^n C_0} + \frac{1}{^n C_1} + \frac{1}{^n C_2} + \dots + \frac{1}{^n C_{n-1}} + \frac{1}{^n C_n}$ So, we can substitute $S_n$ into our equation: $2t_n = n \cdot S_n$

• Step 3: Solve for the ratio. Finally, rearrange the equation to find the required ratio $\frac{t_n}{S_n}$: $\frac{t_n}{S_n} = \frac{n}{2}$

6. Important Tips for Summation Problems

• Recognize Symmetry: Always be on the lookout for symmetry properties in binomial coefficients ($^n C_r = ^n C_{n-r}$) or other sequences. This is often the key to simplification.
• Index Manipulation: Practice rewriting summations by changing the index (e.g., $r \to n-r$). This is a fundamental technique in combinatorics.
• Combine and Conquer: If you have two related sums, or two forms of the same sum, try adding or subtracting them to see if terms cancel or simplify.
• Relate to Known Forms: The goal is often to transform a complex sum into a multiple of a simpler, known sum (like $S_n$ in this case).
• Test with Small Values: If unsure, calculate the first few terms or the sum for small values of $n$ (e.g., $n=0, 1, 2$) to build intuition or verify your derived formula.

7. Summary and Key Takeaway

This problem is a classic example demonstrating the power of the symmetry property of binomial coefficients. By expressing $t_n$ in two equivalent forms (one standard, one transformed using $r \to n-r$ and $^n C_r = ^n C_{n-r}$) and then adding them, we were able to create a new sum where the numerators became constant ($n$). This constant could then be factored out, directly revealing the structure of $S_n$ and leading to a straightforward calculation of the desired ratio $\frac{t_n}{S_n} = \frac{n}{2}$. This technique is widely applicable in problems involving sums of binomial coefficients.