Question

The value of \(\underset{n\rightarrow \infty }{lim}[\frac{1}{1-n^2}+\frac{2}{1-n^2}+\ldots +\frac{n}{1-n^2}]^3\) is

• A. \(8\)
• B. \(-8\)
• C. \(\frac{1}{8}\)
• D. \(\frac{-1}{8}\)

Hint:

Add the numerators with n(n+1)/2, then divide top and bottom by n squared.

Answer 1

The Correct Option is D

Step 1: Approach
Factor the common denominator and simplify using $1-n^2=-(n-1)(n+1)$.

Step 2: Simplify
\[ \frac{n(n+1)}{2}\cdot\frac{1}{-(n-1)(n+1)}=-\frac{n}{2(n-1)} \]

Step 3: Limit
As $n\to\infty$, $\dfrac{n}{n-1}\to1$, so the bracket tends to $-\dfrac12$.

Step 4: Cube
$\left(-\dfrac12\right)^3=-\dfrac18$. The answer is option (D).

Final Answer:
The bracket tends to -1/2, so the limit of its cube is -1/8, option (D). \[ \boxed{-\frac{1}{8}} \]