If $I_n=\int\limits_0^{\pi / 4} \tan ^n x d x$, then $\frac{1}{I_2+I_4}, \frac{1}{I_3+I_5}, \frac{1}{I_4+I_6}, ...$ form
Answer & explanation
Correct answer: option 1
We have,
$I_n+I_{n+2} =\int\limits_0^{\pi / 4} \tan ^n x d x+\int\limits_0^{\pi / 4} \tan ^{n+2} x d x$
$\Rightarrow I_n+I_{n+2} =\int\limits_0^{\pi / 4} \tan ^n x \sec ^2 x d x$
$\Rightarrow I_n+I_{n+2}=\int\limits_0^{\pi / 4} \tan ^n x d(\tan x)=\left[\frac{\tan ^{n+1} x}{n+1}\right]_0^{\pi / 4}=\frac{1}{n+1}$
∴ $\frac{1}{I_n+I_{n+2}}=n+1, n \in N$
$\Rightarrow \frac{1}{I_2+I_4}=3, \frac{1}{I_3+I_5}=4, \frac{1}{I_4+I_6}=5,...$
Clearly, 3, 4, 5, 6, ... are in A.P. with common difference 1.