Functions and functions!

Let's calculate the integral first:- $I_{n} = \int_{0}^{\dfrac{\pi}{4}} tan^{n}xdx$ $I_{n} = \int_{0}^{\dfrac{\pi}{4}} tan^{n-2}x tan^{2}xdx$ $I_{n} = \int_{0}^{\dfrac{\pi}{4}} tan^{n-2}x (sec^{2}x-1)dx$ $I_{n} = \int_{0}^{\dfrac{\pi}{4}}tan^{n-2}x sec^{2}x dx-\int_{0}^{\dfrac{\pi}{4}} tan^{n-2}xdx$ $I_{n} = \Bigg[ \dfrac{tan^{n-1}x}{n-1} \Bigg]_{0}^{\dfrac{\pi}{4}}-I_{n-2}$ $I_{n} + I_{n-2} = \dfrac{1}{n-1} \ldots (1)$ Now we can write that:- $I_{n+2} + I_{n} = \dfrac{1}{n+1}\ldots (2)$ Subtracting $(2)$ from $(1)$ . $I_{n-2}-I_{n+2} = \dfrac{1}{n-1} - \dfrac{1}{n+1}$ $\dfrac{1}{f(n)} = \dfrac{2}{n^{2}-1}$ $f(n) = \dfrac{n^{2}-1}{2}$ Now we can find the sum. $2\sum_{r=2}^{10} f(r) = 2\sum_{r=2}^{10}\dfrac{r^{2}-1}{2}$ $= \sum_{r=2}^{10} (r^{2} -1)$ $= 375$