If $I_n = \int_{0}^{\pi/4} \tan^n x dx$ then $I_{2024} + I_{2026}$ is equal to:

Given: $I_n = \int_{0}^{\pi/4} \tan^n x \, dx$

Need to find: $I_{2024} + I_{2026}$

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Starting with $I_{n+2}$:

$I_{n+2} = \int_{0}^{\pi/4} \tan^{n+2} x \, dx$

$I_{n+2} = \int_{0}^{\pi/4} \tan^n x \cdot \tan^2 x \, dx$

Using the identity $\tan^2 x = \sec^2 x - 1$:

$I_{n+2} = \int_{0}^{\pi/4} \tan^n x (\sec^2 x - 1) \, dx$

$I_{n+2} = \int_{0}^{\pi/4} \tan^n x \sec^2 x \, dx - \int_{0}^{\pi/4} \tan^n x \, dx$

$I_{n+2} = \int_{0}^{\pi/4} \tan^n x \sec^2 x \, dx - I_n$

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For $\int_{0}^{\pi/4} \tan^n x \sec^2 x \, dx$, let $u = \tan x$, so $du = \sec^2 x \, dx$.

When $x = 0$: $u = \tan(0) = 0$

When $x = \pi/4$: $u = \tan(\pi/4) = 1$

The integral becomes:

$\int_{0}^{1} u^n \, du = \left[\frac{u^{n+1}}{n+1}\right]_0^1$

$= \frac{1^{n+1}}{n+1} - \frac{0^{n+1}}{n+1}$

$= \frac{1}{n+1}$

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Substituting back:

$I_{n+2} = \frac{1}{n+1} - I_n$

Therefore:

$I_n + I_{n+2} = \frac{1}{n+1}$

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Using the formula with $n = 2024$:

$I_{2024} + I_{2026} = \frac{1}{2024+1}$

$= \frac{1}{2025}$

Therefore, $I_{2024} + I_{2026} = \dfrac{1}{2025}$