Question

The sum to 'n' terms of the series
$ta{n}^{-1}\left(\frac{1}{2}\right)+ta{n}^{-1}\left(\frac{2}{9}\right)+ta{n}^{-1}\left(\frac{1}{8}\right)+ta{n}^{-1}\left(\frac{2}{25}\right)+ta{n}^{-1}\left(\frac{1}{18}\right)+.....\infty$ terms

• A. $ta{n}^{-1}3$
• B. $co{t}^{-1}\left(\frac{1}{3}\right)$
• C. $ta{n}^{-1}\left(\frac{1}{3}\right)$
• D. None of these

Answer 1

Answer: (A), (B)

The correct options are
A

$ta{n}^{-1}3$

B

$co{t}^{-1}\left(\frac{1}{3}\right)$

$T_r=ta{n}^{-1}{\left(\frac{\sqrt{2}}{r+1}\right)}^2$

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

$=ta{n}^{-1}\frac{r+2-r}{1+r(r+2)}=ta{n}^{-1}(r+2)-ta{n}^{-1}r$

$S_{\infty }=-ta{n}^{-1}2-ta{n}^{-1}1$

$\therefore S_{\infty }=ta{n}^{-1}\left(3\right)=co{t}^{-1}\left(\frac{1}{3}\right)$