Q103 / 160 · AP EAMCET 2018 (23 Apr Shift 2) · Mathematics
$\lim _{n \rightarrow-\infty} \frac{\left[6^2+12^2+18^2+\ldots .+(6 n)^2\right]^2}{[5+10+15+\ldots .+5 n]\left[2^3+4^3+6^3+\ldots .+8 n^3\right]}=$
- A$\frac{4}{5}$
- Correct answerB$\frac{144}{5}$
- C$\frac{4}{25}$
- D$\frac{144}{25}$
Solution
$\begin{aligned} & \text { } \lim _{n \rightarrow \infty} \frac{\left[6^2+12^2+18^2+\ldots+(6 n)^2\right]^2}{[5+10+15+\ldots+5 n]\left[2^3+4^3+6^3+\ldots+8 n^3\right]} \\ & =\lim _{n \rightarrow \infty} \frac{6^4\left[1^2+2^2+3^2+\ldots+n^2\right]^2}{(5)[1+2+3+\ldots+n]\left(2^3\right)} \\ & =\lim _{n \rightarrow \infty} \frac{6^4\left(\frac{n(n+1)(2 n+1)}{6}\right)^2}{5\left(\frac{n(n+1)}{2}\right) \times 2^3 \times\left(\frac{n(n+1)}{2}\right)^2} \\ & \left.=\lim _{n \rightarrow \infty} \frac{\left(\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right)\right)^3}{6}\right)^2 \times\left(\frac{\left.1+\frac{1}{n}\right) \times 2^3\left(\frac{1+\frac{1}{n}}{2}\right)^2}{2}\right) \\ & =\frac{5 \times\left(\frac{1}{6}\right)^4 \times \frac{1}{6^2} \times 2^2}{5 \times \frac{1}{2} \times 2^3 \times \frac{1}{2^2}}=\frac{144}{5} .\end{aligned}$
Hence, the correct answer is option: 'B'.