Question:

Step 1: The problem asks us to find the value of the given sum $\dfrac{{}^{100}C_{50}}{51}+\dfrac{{}^{100}C_{51}}{52}+\cdots+\dfrac{{}^{100}C_{100}}{101}$. To approach this, we need to first understand the properties of combinations, denoted by ${}^{n}C_{r}$, which represents the number of ways to choose $r$ items from a set of $n$ distinct items. Step 2: We can use the property of combinations that states ${}^{n}C_{r} = {}^{n}C_{n-r}$. This property will be useful in simplifying the given sum. Additionally, we know that the sum of combinations of the form ${}^{n}C_{r}$ for $r$ ranging from $0$ to $n$ is $2^{n}$, which is a fundamental property related to the binomial theorem. Step 3: Now, let's consider the given sum and try to simplify it using the properties mentioned above. We have $\dfrac{{}^{100}C_{50}}{51}+\dfrac{{}^{100}C_{51}}{52}+\cdots+\dfrac{{}^{100}C_{100}}{101}$. Using the property ${}^{n}C_{r} = {}^{n}C_{n-r}$, we can rewrite the sum as $\dfrac{{}^{100}C_{50}}{51}+\dfrac{{}^{100}C_{49}}{52}+\cdots+\dfrac{{}^{100}C_{0}}{101}$. Step 4: To further simplify the sum, we can use the fact that $\dfrac{{}^{n}C_{r}}{n+1-r} = \dfrac{{}^{n}C_{r}}{n+1-r} \cdot \dfrac{n+1-r}{n+1} \cdot \dfrac{n+1}{n+1} = \dfrac{1}{n+1} \cdot \dfrac{{}^{n+1}C_{r}}{n+1-r}$. Applying this to our sum, we get $\dfrac{1}{101} \cdot ({}^{101}C_{51} + {}^{101}C_{52} + \cdots + {}^{101}C_{101})$. Step 5: Now, recall that the sum of combinations of the form ${}^{n}C_{r}$ for $r$ ranging from $0$ to $n$ is $2^{n}$. Therefore, the sum ${}^{101}C_{51} + {}^{101}C_{52} + \cdots + {}^{101}C_{101}$ can be rewritten as $2^{100}$, since we are considering the sum of combinations from ${}^{101}C_{0}$ to ${}^{101}C_{101}$ and then subtracting the sum of combinations from ${}^{101}C_{0}$ to ${}^{101}C_{50}$, which equals $2^{100}$. Step 6: Substituting this result back into our expression, we get $\dfrac{1}{101} \cdot 2^{100}$. Therefore, the value of the given sum is $\dfrac{2^{100}}{101}$. Step 7: Comparing this result with the given options, we find that the correct answer is Option 4: $\dfrac{2^{100}}{101}$. Hence, the final answer is $\boxed{4}$, which corresponds to the value $\dfrac{2^{100}}{101}$.