The value of $${{{e^{ - {\pi \over 4}}} + \int\limits_0^{{\pi \over 4}} {{e^{ - x}}{{\tan }^{50}}xdx} } \over {\int\limits_0^{{\pi \over 4}} {{e^{ - x}}({{\tan }^{49}}x + {{\tan }^{51}}x)dx} }}$$ is

Among

(S1): $$\lim_\limits{n \rightarrow \infty} \frac{1}{n^{2}}(2+4+6+\ldots \ldots+2 n)=1$$

(S2) : $$\lim_\limits{n \rightarrow \infty} \frac{1}{n^{16}}\left(1^{15}+2^{15}+3^{15}+\ldots \ldots+n^{15}\right)=\frac{1}{16}$$

$$\int_\limits{0}^{\infty} \frac{6}{e^{3 x}+6 e^{2 x}+11 e^{x}+6} d x=$$

If $$f: \mathbb{R} \rightarrow \mathbb{R}$$ be a continuous function satisfying $$\int_\limits{0}^{\frac{\pi}{2}} f(\sin 2 x) \sin x d x+\alpha \int_\limits{0}^{\frac{\pi}{4}} f(\cos 2 x) \cos x d x=0$$, then the value of $$\alpha$$ is :