Q58 / 150 · MHT CET 2026 (20 April Shift 2) · Mathematics
The value of $\lim\limits_{n \to \infty}\left[\dfrac{1}{1 - n^2} + \dfrac{2}{1 - n^2} + \ldots + \dfrac{n}{1 - n^2}\right]^3$ is
- A$8$
- B$-8$
- C$\dfrac{1}{8}$
- Correct answerD$\dfrac{-1}{8}$
Solution
Let $L = \lim\limits_{n \to \infty}\left[\dfrac{1}{1 - n^2} + \dfrac{2}{1 - n^2} + \ldots + \dfrac{n}{1 - n^2}\right]^3$
The expression inside the bracket can be simplified by taking the common denominator:
$S_n = \dfrac{1 + 2 + \ldots + n}{1 - n^2}$
Using the sum of the first $n$ natural numbers, $1 + 2 + \ldots + n = \dfrac{n(n+1)}{2}$:
$S_n = \dfrac{\dfrac{n(n+1)}{2}}{1 - n^2}$
$S_n = \dfrac{n(n+1)}{2(1 - n)(1 + n)}$
$S_n = \dfrac{n}{2(1 - n)}$
Taking the limit as $n \to \infty$:
$\lim\limits_{n \to \infty} S_n = \lim\limits_{n \to \infty} \dfrac{n}{2(1 - n)} = \lim\limits_{n \to \infty} \dfrac{1}{2\left(\dfrac{1}{n} - 1\right)} = -\dfrac{1}{2}$
Therefore, the required limit is:
$L = \left(-\dfrac{1}{2}\right)^3 = -\dfrac{1}{8}$
Answer: $\dfrac{-1}{8}$
The expression inside the bracket can be simplified by taking the common denominator:
$S_n = \dfrac{1 + 2 + \ldots + n}{1 - n^2}$
Using the sum of the first $n$ natural numbers, $1 + 2 + \ldots + n = \dfrac{n(n+1)}{2}$:
$S_n = \dfrac{\dfrac{n(n+1)}{2}}{1 - n^2}$
$S_n = \dfrac{n(n+1)}{2(1 - n)(1 + n)}$
$S_n = \dfrac{n}{2(1 - n)}$
Taking the limit as $n \to \infty$:
$\lim\limits_{n \to \infty} S_n = \lim\limits_{n \to \infty} \dfrac{n}{2(1 - n)} = \lim\limits_{n \to \infty} \dfrac{1}{2\left(\dfrac{1}{n} - 1\right)} = -\dfrac{1}{2}$
Therefore, the required limit is:
$L = \left(-\dfrac{1}{2}\right)^3 = -\dfrac{1}{8}$
Answer: $\dfrac{-1}{8}$
Hence, the correct answer is option: 'D'.