Question

For $ x \in\left(0, \frac{\pi}{4}\right) $
$R _{ n }=\displaystyle\sum_{ r =1}^{2 n } \sin \left(\sin ^{-1} x ^{4 r -3}\right), \,\,\, S _{ n }=\displaystyle\sum_{ r =1}^{2 n } \cos \left(\cos ^{-1} x ^{4 r -2}\right), $
$T _{ n }=\displaystyle\sum_{ r =1}^{2 n } \tan \left(\tan ^{-1} x ^{4 r -1}\right), \,\,\, U _{ n }=\displaystyle\sum_{ r =1}^{2 n } \cot \left(\cot ^{-1} x ^{4 r }\right)$
where $n \in N$ and $n \geq 4$

• A. $R _{ n }< S _{ n }< T _{ n }< U _{ n }$
• B. $R_n>S_n>T_n >U_n$
• C. $\underset{n \rightarrow \infty}{\text{Lim}} \left(R_n+S_n+T_n+U_n\right)$ is equal to $\frac{x}{1-x}$
• D. The value of $x$ for which $R_n+S_n=T_n+U_n$ is $2 \sin \frac{\pi}{10}$

Answer 1

Answer: (B), (C)

$\Theta R_n=\displaystyle\sum_{r=1}^{2 n} x^{4 r-3}=x+x^5+x^9 \ldots \ldots \ldots .$.
$S_n=\displaystyle\sum_{r=1}^{2 n} x^{4 r-2}=x^2+x^6+x^{10} \ldots \ldots \ldots .$
$T_n=\displaystyle\sum_{r=1}^{2 n} x^{4 r-1}=x^3+x^7+x^{11} \ldots \ldots \ldots . $
$U_n=\displaystyle\sum_{r=1}^{2 n} x^{4-r}=x^4+x^8+x^{12} \ldots \ldots \ldots . $
clearly, $R _{ n }> S _{ n }> T _{ n }> U _{ n } $ as $x \in\left(0, \frac{\pi}{4}\right)$
and $ \underset{n \rightarrow \infty}{\text{Lim}} \left(R_n+S_n+T_n+U_n\right)=x+x^2+\ldots \ldots \ldots \ldots . .=\frac{x}{1-x}$
$\Theta R_n+S_n=T_n+U_n $
$\Rightarrow \frac{x\left(1-x^{8 n}\right)}{1-x^4}+\frac{x^2\left(1-x^{8 n}\right)}{1-x^4}=\frac{x^3\left(1-x^{8 n}\right)}{1-x^4}+\frac{x^4\left(1-x^{8 n}\right)}{1-x^4} $
$ \Rightarrow x+x^2=x^3+x^4 \Rightarrow \left(x+x^2\right)\left(x^2-1\right)=0 \Rightarrow \text { no value of } x$