KVPY #20

$\begin{aligned} I_n & = \int_0^\frac \pi 4 \tan^n x \ dx & \small \color{#3D99F6} \text{Let }t = \tan x, \ dt = \sec^2 x \ dx = 1+t^2 \ dx \\ & = \int_0^1 \frac {t^n}{1+t^2} dt \end{aligned}$

$\begin{aligned} \implies I_n + I_{n+2} & = \int_0^1 \frac {t^n}{1+t^2} dt + \int_0^1 \frac {t^{n+2}}{1+t^2} dt \\ & = \int_0^1 \frac {t^n+t^{n+2}}{1+t^2} dt \\ & = \int_0^1 t^n \ dt \\ & = \frac 1{n+1} \end{aligned}$

$\implies \dfrac 1{I_2+I_4} = \frac 1{\frac 12} = 2$ , $\dfrac 1{I_3+I_5} = 3$ , and $\dfrac 1{I_4+I_6} = 4$ . They are in an arithmetic progression .