Question:

Step 1: Understand the problem and the expression for $P_n$. We are given the expression $P_n=C_0-C_1+\dfrac{2^2}{3}C_2-\dfrac{2^3}{4}C_3+\cdots+\dfrac{(-2)^n}{n+1}C_n$, where $C_r$ denotes the coefficient of $x^r$ in the binomial expansion of $(1+x)^n$. Our goal is to find the value of $\displaystyle\sum_{n=1}^{25}\dfrac{1}{P_{2n}}$. Step 2: Find the relationship between $P_n$ and $n$ to simplify the calculation of $\dfrac{1}{P_{2n}}$. To simplify the calculation, we need to find a relationship between $P_n$ and $n$. By analyzing the given expression for $P_n$, it can be shown that $\dfrac{1}{P_{2n}} = 2n + 1$. This step involves understanding the properties of binomial coefficients and their relation to the given series. Step 3: Calculate the sum $\displaystyle\sum_{n=1}^{25}\dfrac{1}{P_{2n}}$ using the derived relationship. Using the relationship $\dfrac{1}{P_{2n}} = 2n + 1$, we can now calculate the sum $\displaystyle\sum_{n=1}^{25}\dfrac{1}{P_{2n}}$. This sum becomes $\sum_{n=1}^{25} (2n + 1)$. Step 4: Evaluate the sum $\sum_{n=1}^{25} (2n + 1)$. The sum $\sum_{n=1}^{25} (2n + 1)$ can be evaluated as $2\sum_{n=1}^{25} n + \sum_{n=1}^{25} 1$. Using the formula for the sum of the first $n$ natural numbers, $\sum_{n=1}^{N} n = \dfrac{N(N+1)}{2}$, we can calculate this sum. Step 5: Apply the formula for the sum of the first $n$ natural numbers. Applying the formula, we get $2\sum_{n=1}^{25} n + \sum_{n=1}^{25} 1 = 2 \cdot \dfrac{25 \cdot 26}{2} + 25 = 2 \cdot 325 + 25 = 650 + 25 = 675$. Step 6: Conclude the final answer. The final answer is: $\boxed{675}$