Question

The sum of n terms of the series $co{t}^{-1}3+co{t}^{-1}7+co{t}^{-1}13+co{t}^{-1}21+......$ is

• A. $ta{n}^{-1}\frac{n-1}{n+2}$
• B. $ta{n}^{-1}\frac{n}{n+2}$
• C. $ta{n}^{-1}\frac{n+1}{n+2}$
• D. none of these

Answer 1

The correct option is B $ta{n}^{-1}\frac{n}{n+2}$

$co{t}^{-1}3+co{t}^{-1}7+co{t}^{-1}13+co{t}^{-1}21+.....$

$ta{n}^{-1}\frac{1}{3}+ta{n}^{-1}\frac{1}{7}+ta{n}^{-1}\frac{1}{13}+ta{n}^{-1}\frac{1}{21}+.......$

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

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

$\left(\right(ta{n}^{-1}2-ta{n}^{-1}1)+(ta{n}^{-1}3-ta{n}^{-1}2)+(ta{n}^{-1}4-ta{n}^{-1}3)and......+(ta{n}^{-1}(n+1)-ta{n}^{-1}n\left)\right)$

$=ta{n}^{-1}(n+1)-ta{n}^{-1}1$

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