MHTCET 10 May 2024 Shift 1 Paper

Concept:Use standard formulas for $\sum r$, $\sum r^2$ and $\sum r^3$, then evaluate the limit by comparing the highest degree terms in the numerator and denominator.Explanation:Given:$S_1 = \sum_{r=1}^{n} r = \frac{n(n+1)}{2}$$S_2 = \sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6}$$S_3 = \sum_{r=1}^{n} r^3 = \left(\frac{n(n+1)}{2}\right)^2$Substitute these into the required limit:$\lim_{n \to \infty} \frac{S_1\left(1+\frac{S_3}{4}\right)}{S_2^2} < br>= \lim_{n \to \infty} \frac{\frac{n(n+1)}{2}\left(1+\frac{n^2(n+1)^2}{16}\right)}{\frac{n^2(n+1)^2(2n+1)^2}{36}}$Simplify the fraction:$= \lim_{n \to \infty} \frac{n(n+1)}{2}\left(1+\frac{n^2(n+1)^2}{16}\right) \times \frac{36}{n^2(n+1)^2(2n+1)^2}$$= \lim_{n \to \infty} 18 \cdot \frac{1+\frac{n^2(n+1)^2}{16}}{n(n+1)(2n+1)^2}$For large $n$, the highest degree term in the numerator is $\frac{n^4}{16}$, and in the denominator it is $4n^4$.Therefore, the limit becomes:$18 \times \frac{\frac{n^4}{16}}{4n^4} = 18 \times \frac{1}{64} = \frac{9}{32}$Answer:$\frac{9}{32}$, which is option C.