For non-negative integers n, let f(n) = ∑ _ k = 0 ^ n ((k + 1)/(n +...

$$\begin{array}{ll}f\left(n\right) = \dfrac{\mathop{∑}_{k = 0}^{n}\sin \left(\left(\dfrac{k + 1}{n + 2}\right)\pi\right)\sin \left(\left(\dfrac{k + 2}{n + 2}\right)\pi\right)}{\mathop{∑}_{k = 0}^{n}si{n}^{2}\left(\left(\dfrac{k + 1}{n + 2}\right)\pi\right)}\end{array}$$

\left($$\begin{array}{ll}∵ 2sinC sin D = \cos \left(C - D\right) - \cos \left(C + D\right) \\ ∵ 2\sin^{2} \theta = 1 - \cos 2\theta\end{array}$$\right)$<br />$f\left(n\right) = \dfrac{\mathop{∑}_{k = 0}^{n}\left(\cos \left(\dfrac{\pi}{n + 2}\right) - \cos \left(\dfrac{2k + 3}{n + 2}\pi\right)\right)}{\mathop{∑}_{k = 0}^{n}\left(1 - \cos \left(\left(\dfrac{2k + 2}{n + 2}\right)\pi\right)\right)}$<br />$f\left(n\right) = \dfrac{\left(n + 1\right)\cos \left(\dfrac{\pi}{n + 2}\right) - \mathop{∑}_{k = 0}^{n}\cos \left(\dfrac{2k + 3}{n + 2}\right)\pi}{\left(n + 1\right) - \mathop{∑}_{k = 0}^{n}\cos \left(\dfrac{2k + 2}{n + 2}\right)\pi}$<br />Now$cos \alpha + \cos \left(\alpha + \beta\right) + \ldots + \cos \left(\alpha + \left(n - 1\right)\beta\right) = \dfrac{\sin \left(\dfrac{n\beta}{2}\right)}{\sin \left(\dfrac{\beta}{2}\right)}\cos \left(\alpha + \dfrac{\left(n - 1\right)\beta}{2}\right)$<br />$f\left(n\right) = \dfrac{\left(n + 1\right)\cos \left(\dfrac{\pi}{n + 2}\right) - \dfrac{\sin \left(\dfrac{n + 1}{n + 2}\pi\right)}{\sin \left(\dfrac{\pi}{n + 2}\right)}\cos \left(\dfrac{n + 3}{n + 2}\pi\right)}{\left(n + 1\right) - \dfrac{\sin \left(\dfrac{n + 1}{n + 2}\right)\pi}{\sin \left(\dfrac{\pi}{n + 2}\right)}\cos \left(\pi\right)}$<br />In numerator cosine sum$\alpha = \dfrac{3\pi}{n + 2}$<br />$\beta = \dfrac{2\pi}{n + 2}$<br />Number of terms$= \left(n + 1\right)$<br />In denominator cosine sum$\alpha = \dfrac{2\pi}{n + 2}$<br />$\beta = \dfrac{2\pi}{n + 2}$<br />Number of terms$= \left(n + 1\right)$<br />Now$\sin \left(\dfrac{n + 1}{n + 2}\pi\right) = \sin \left(\pi - \dfrac{\pi}{n + 2}\right) = sin \dfrac{\pi}{n + 2}$<br />$\cos \left(\dfrac{n + 3}{n + 2}\pi\right) = \cos \left(\pi + \dfrac{\pi}{n + 2}\right) = - cos \dfrac{\pi}{n + 2}$<br />$f\left(n\right) = \dfrac{\left(n + 1\right)\cos \left(\dfrac{\pi}{n + 2}\right) + \cos \left(\dfrac{\pi}{n + 2}\right)}{\left(n + 1\right) - \left(- 1\right)}$<br />$f\left(n\right) = \dfrac{\cos \left(\dfrac{\pi}{n + 2}\right)\left(n + 2\right)}{\left(n + 2\right)}$<br />$f\left(n\right) = \cos \left(\dfrac{\pi}{n + 2}\right)$<br />Hence, \left(A\right)f$\left(4\right) = cos $\left(\dfrac{\pi}{6}\right) = \dfrac{\sqrt{3}}{2}$
\left(B\right)\lim_{n + \infty} f\left(n\right) = \cos \left(0\right) = 1$<br />\left(C\right)f\left(6\right) = \cos \left(\dfrac{\pi}{8}\right):$ then $\alpha = \tan \left(\dfrac{\pi}{8}\right) = \sqrt{2} - 1$
$\alpha = \sqrt{2}\left(5\right) = \cos \left(\dfrac{\pi}{7}\right)$
$\sin \left(7\cos^{- 1} \left(cos \dfrac{\pi}{7}\right)\right) = sin \pi = 0$