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${S_n} = \frac{1}{3} + \frac{1}{{15}} + \frac{1}{{35}} + \frac{1}{{63}} + ......... = ?$

${t_r} = \frac{1}{{4{r^2} - 1}}$

$ = \frac{1}{{(2r - 1)(2r + 1)}} = \frac{1}{2}\left( {\frac{1}{{2r - 1}} - \frac{1}{{2r + 1}}} \right)$

${S_n} = \sum\limits_{r = 1}^n {\frac{1}{2}\left( {\frac{1}{{2r - 1}} - \frac{1}{{2r + 1}}} \right)} $

${S_n} = \frac{1}{2}\left[ {\left( {\frac{1}{1} - \frac{1}{3}} \right) + \left( {\frac{1}{3} - \frac{1}{5}} \right) + \left( {\frac{1}{5} - \frac{1}{7}} \right) + ....... + \left( {\frac{1}{{2n - 1}} - \frac{1}{{2n + 1}}} \right)} \right]$

${S_n} = \frac{1}{2}\left[ {1 - \frac{1}{{2n + 1}}} \right] = \frac{n}{{2n + 1}}$


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${\log _{\sqrt 5 }}\left[ {3 + \cos \left( {\frac{{3\pi }}{4} + x} \right) + \cos \left( {\frac{\pi }{4} + x} \right) + \cos \left( {\frac{\pi }{4} - x} \right) - \cos \left( {\frac{{3\pi }}{4} - x} \right)} \right]$

The range of the function $f(x) = {\log _{\sqrt 5 }}\left[ {3 + \cos \left( {\frac{{3\pi }}{4} + x} \right) + \cos \left( {\frac{\pi }{4} + x} \right) + \cos \left( {\frac{\pi }{4} - x} \right) - \cos \left( {\frac{{3\pi }}{4} - x} \right)} \right]$ is: (A) $[ - 2,2]$ (B) $\left[ {\frac{1}{{\sqrt 5 }},\sqrt 5 } \right]$ (C) $(0,\sqrt 5 )$ (D) $[ 0,2]$ Solution We have, $f(x) = {\log _{\sqrt 5 }}\left( {3 - 2\sin \frac{{3\pi }}{4}\sin x + 2\cos \frac{\pi }{4}\cos x} \right)$ $ \Rightarrow f(x) = {\log _{\sqrt 5 }}\left[ {3 + \sqrt 2 (\cos x - \sin x)} \right]$ Now, $ - \sqrt 2  \le \cos x - \sin x \le \sqrt 2 $ $\therefore - 2 \le \sqrt 2 (\cos x - \sin x) \le 2$ $\therefore 1 \le 3 + \sqrt 2 (\cos x - \sin x) \le 5$ $\therefore{\log _{\sqrt 5 }}1 \le {\log _{\sqrt 5 }}[3 + \sqrt 2 (\cos x - \sin x)] \le {\log _{\sqrt 5 }}5$ $ \Rightarrow 0 \le f(x) \le 2$ Answer: (D)
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