Question

$\frac{1}{n^{2}} + \frac{1}{2n^{4}} + \frac{1}{3n^{6}} + ......\infty =$

• A. $\log_{e}\left( \frac{n^{2}}{n^{2} + 1} \right)$
• B. $\log_{e}\left( \frac{n^{2} + 1}{n^{2}} \right)$
• C. $\log_{e}\left( \frac{n^{2}}{n^{2} - 1} \right)$
• D. None of these

Answer 1

Answer: (C)

$\log_{e}\left( \frac{n^{2}}{n^{2} - 1} \right)$

Answer 2

$e^{a - bx} - 1$

$1 + a\log_{e}(a - bx)$.

Trick : Putting$e^{- bx}$the sum of the series upto 4 terms is $y = - \left( x^{3} + \frac{x^{6}}{2} + \frac{x^{9}}{3} + ..... \right)$... and option (1)

= – 0.223...., (2) =0.223....., (3) =0.2876.....

Hence answer is (3).