Let $$C_{r}$$ denote the coefficient of $$x^{r}$$ in the binomial expansion of $$(1+x)^{n}, n\in N, 0\leq r\leq n$$. If $$P_{n}= C_{0}-C_{1}+\frac{2^{2}}{3}C_{2}-\frac{2^{3}}{4}C_{3}+.....+\frac{(-2)^{n}}{n+1}C_{n}, \text{then the value of} \sum_{n=1}^{25} \frac{1}{P_{2n}} $$ equals.

$$P_n = \int_0^1 (1-2x)^n dx$$. Let $$u=1-2x$$: $$= \frac{-1}{2}\int_1^{-1}u^n du = \frac{1}{2}\frac{u^{n+1}}{n+1}\Big|_{-1}^{1} = \frac{1-(-1)^{n+1}}{2(n+1)}$$.

For $$n$$ even: $$P_{2n} = \frac{1-(-1)^{2n+1}}{2(2n+1)} = \frac{2}{2(2n+1)} = \frac{1}{2n+1}$$.

$$\sum_{n=1}^{25}\frac{1}{P_{2n}} = \sum_{n=1}^{25}(2n+1) = \sum_{n=1}^{25}(2n+1) = 3+5+...+51 = 25 \times 27 = 675$$.

The answer is Option 4: 675.