MHTCET 20 Apr 2026 Shift 2 Paper

Concept:The sum of the first $n$ natural numbers helps simplify the bracket, followed by evaluating the limit as $n \to \infty$.Explanation:The given bracket is$$\frac{1}{1-n^{2}}+\frac{2}{1-n^{2}}+\cdots+\frac{n}{1-n^{2}}$$Since the denominator is common, add the numerators:$$\frac{1+2+\cdots+n}{1-n^{2}}$$Use the formula for the sum of the first $n$ natural numbers:$$1+2+\cdots+n=\frac{n(n+1)}{2}$$So the bracket becomes$$\frac{\frac{n(n+1)}{2}}{1-n^{2}}$$Factorise $1-n^{2}$:$$1-n^{2}=(1-n)(1+n)$$Thus, the expression is$$\frac{n(n+1)}{2(1-n)(1+n)}$$Cancel $(n+1)$ with $(1+n)$:$$\frac{n}{2(1-n)}$$Rewrite it as$$\frac{n}{2(1-n)}=-\frac{n}{2(n-1)}$$Now take the limit as $n \to \infty$:$$-\frac{n}{2(n-1)} \to -\frac{1}{2}$$The required limit is the cube of this value:$$\left(-\frac{1}{2}\right)^{3}=-\frac{1}{8}$$Hence, the correct option is Option D $\frac{-1}{8}$.Answer:$$-\frac{1}{8}$$Option D.