Let ${\mathrm{I}}_{\mathrm{n}}={\int }_0^{\mathrm{\pi }/4} {\tan }^{\mathrm{n}}\mathrm{xdx},\text{ then }\frac{1}{{\mathrm{I}}_2+{\mathrm{I}}_4},\frac{1}{{\mathrm{I}}_3+{\mathrm{I}}_5},\frac{1}{{\mathrm{I}}_4+{\mathrm{I}}_6},\frac{1}{{\mathrm{I}}_5+{\mathrm{I}}_7}$, form an :

$$\begin{gathered} \begin{array}{l}\text{ In }={\int }_0^{\mathrm{\pi }/4} {\tan }^{\mathrm{n}}\mathrm{x}={\int }_0^{\mathrm{\pi }/4} {\tan }^2\mathrm{x}\cdot {\tan }^{\mathrm{n}-2}\mathrm{xdx}\\ ={\int }_0^{\mathrm{\pi }/4} {\sec }^2{\mathrm{xtan}}^{n-2}\mathrm{xdx}-{\int }_0^{\mathrm{\pi }/4} {\tan }^{\mathrm{n}-2}\mathrm{xdx}\\ ={\left.\frac{{\tan }^{\mathrm{n}-1}\mathrm{x}}{\mathrm{n}-1}\right|}_0^{\mathrm{\pi }/4}-{\mathrm{I}}_{\mathrm{n}-2}\\ {\mathrm{I}}_{\mathrm{n}}+{\mathrm{I}}_{\mathrm{n}-2}=\frac{1}{\mathrm{n}-1} \\ \therefore \frac{1}{{\mathrm{I}}_2+{\mathrm{I}}_4}=3\\ \frac{1}{{\mathrm{I}}_3+{\mathrm{I}}_5}=4\\ \frac{1}{{\mathrm{I}}_4+{\mathrm{I}}_6}=5\end{array} \end{gathered}$$