Animated Solution for Mathematics - Binomial Theorem: If $S_n={\sum }_{r=0}^n\frac{1}{{}^n{C}_r}$ and $t_n={\sum }_{r=0}^n\frac{r}{{}^n{C}_r}$, then $\frac{t_n}{S_n}$ is equal to

We are tasked with evaluating the relationship between two sums involving binomial coefficients: $S_n={\sum }_{r=0}^n\frac{1}{{}^n{C}_r}$ and $t_n={\sum }_{r=0}^n\frac{r}{{}^n{C}_r}$.

While expanding the binomial coefficients is a valid algebraic approach, it is often computationally heavy. In JEE Advanced mathematics, identifying structural symmetry is a far more efficient strategy.

The core of this problem lies in the fundamental property of binomial coefficients:

Think of this as a mirror. If you examine any row of Pascal's triangle, it reads the same from left to right as it does from right to left. This is not merely a coincidence; it is a structural reality of how we choose items from a set.

Consider the sum $t_n={\sum }_{r=0}^n\frac{r}{{}^n{C}_r}$. We can utilize the property that the sum of a function over a range remains invariant if we reverse the order of summation.

By replacing $r$ with $n-r$, the expression becomes:

Applying the symmetry property ${}^n{C}_{n-r}={}^n{C}_r$ to the denominator, we transform the expression into:

We can now split this summation into two distinct parts:

Observe that the first part is simply $n$ times our original sum $S_n$, while the second part is exactly the original $t_n$ we started with. This yields the elegant equation:

Rearranging the terms, we get $2t_n=n{S}_n$. Dividing both sides by $2S_n$, we arrive at the final result:

This problem demonstrates that the most complex-looking expressions often yield to simple, elegant symmetry. Always keep this reversal trick in your toolkit—it is a powerful strategy for solving many JEE-level problems.