We have $P_n=\sum_{r=0}^{n}\dfrac{(-2)^r}{r+1}C_r$. Using $\sum_{r=0}^{n} C_r\dfrac{a^{r+1}}{r+1}=\dfrac{(1+a)^{n+1}-1}{n+1}$ with $a=-2$, we get $P_n=-\dfrac12\cdot\dfrac{(-1)^{n+1}-1}{n+1}$. For even $n=2m$, $(-1)^{2m+1}=-1$, so $P_{2m}=\dfrac{1}{2m+1}$. Therefore $\dfrac{1}{P_{2n}}=2n+1$. Hence $\sum_{n=1}^{25}\dfrac{1}{P_{2n}}=\sum_{n=1}^{25}(2n+1)=2\cdot\dfrac{25\cdot26}{2}+25=650+25=675$.