$\lim _{n \rightarrow \infty} \frac{2^2+4^2+6^2+\ldots+(2 n)^2}{n^3}=$

Question

Aicharya · Accepted Answer

$\begin{aligned} & \lim _{n \rightarrow \infty} \frac{2^2+4^2+6^2+\ldots+(2 n)^2}{n^3} \ & =4 \lim _{n \rightarrow \infty} \frac{1^2+2^2+3^2+\ldots+n^2}{n^3} \ & =4 \lim _{n \rightarrow \infty} \frac{n(n+1)(2 n+1)}{6 n^3}=4 \lim _{n \rightarrow \infty} \frac{1 \cdot\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right)}{6} \ & =4 \times \frac{2}{6}=\frac{4}{3} \end{aligned}$ Hence, option (b) is correct.

\(\lim _{n \rightarrow \infty} \frac{2^2+4^2+6^2+\ldots+(2 n)^2}{n^3}=\)

Solution