Correct Answer - Option (B)
let $L = \underset{n \to \infty}{\lim}\bigg( \frac{1}{1-n^2} + \frac{2}{1-n^2} +\frac{3}{1-n^2} + \frac{4}{1-n^2} + ... + \frac{n}{1-n^2}\bigg) $
taking $(1 - n^2)$ common,
$L = \underset{n \to \infty}{\lim}\bigg( \frac{1}{1-n^2} \cdot (1 + 2 + 3 + ...+n) \bigg) $
Now, sum of first $n$ natural numbers is $\frac{n(n+1)}{2}$ so substituting,
$L = \underset{x \to \infty}{\lim} \bigg( \frac{n(n+1)}{2} \cdot \frac{1}{(1-n)(1+n)}\bigg) = \frac{1}{2} \cdot \underset{x \to \infty}{\lim} \bigg( \frac{n}{1-n} \bigg)$
By L' Hopital's rule,
$\underset{x \to \infty}{\lim} \bigg( \frac{n}{1-n} \bigg) = \underset{x \to \infty}{\lim} \bigg( \frac{1}{-1} \bigg) = -1$
Hence, $L = - \frac{1}{2}$