Question:

Step 1: Express the given series in a more general form and identify a common factor. Let the given series be $S$. We can observe that each term is of the form $\dfrac{1}{m!(n-m)!}$ where $n=51$ and $m$ takes odd values. To transform these terms into binomial coefficients, we can multiply the entire series by $51!$. $$S = \dfrac{1}{1!50!}+\dfrac{1}{3!48!}+\dfrac{1}{5!46!}+\cdots+\dfrac{1}{49!2!}+\dfrac{1}{51!1!}$$ Step 2: Multiply the series by $51!$ to express terms as binomial coefficients. Multiplying both sides by $51!$ will convert each term $\dfrac{1}{k!(51-k)!}$ into $\dfrac{51!}{k!(51-k)!}$, which is the binomial coefficient $\binom{51}{k}$. $$51!S = \dfrac{51!}{1!50!}+\dfrac{51!}{3!48!}+\dfrac{51!}{5!46!}+\cdots+\dfrac{51!}{49!2!}+\dfrac{51!}{51!1!}$$ Step 3: Rewrite the series using binomial coefficient notation. Using the definition $\binom{n}{k} = \dfrac{n!}{k!(n-k)!}$, we can rewrite the expression obtained in Step 2. $$51!S = \binom{51}{1} + \binom{51}{3} + \binom{51}{5} + \cdots + \binom{51}{49} + \binom{51}{51}$$ Step 4: Apply the identity for the sum of odd-indexed binomial coefficients. We know that for any positive integer $n$, the sum of odd-indexed binomial coefficients is equal to $2^{n-1}$. That is, $\binom{n}{1} + \binom{n}{3} + \binom{n}{5} + \cdots = 2^{n-1}$. In this case, $n=51$. $$\binom{51}{1} + \binom{51}{3} + \binom{51}{5} + \cdots + \binom{51}{49} + \binom{51}{51} = 2^{51-1}$$ $$ = 2^{50}$$ Step 5: Solve for the value of the original series $S$. Substitute the value of the sum from Step 4 back into the equation from Step 3. $$51!S = 2^{50}$$ $$S = \dfrac{2^{50}}{51!}$$ The final answer is $\dfrac{2^{50}}{51!}$. The final answer matches Option 2.