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\(\lim _{n \rightarrow \infty} \frac{2^2+4^2+6^2+\ldots+(2 n)^2}{n^3}=\)
Options:
Solution:
2042 Upvotes
Verified Answer
The correct answer is:
\(\frac{4}{3}\)
\(\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{2^2+4^2+6^2+\ldots+(2 n)^2}{n^3} \\
& =4 \lim _{n \rightarrow \infty} \frac{1^2+2^2+3^2+\ldots+n^2}{n^3} \\
& =4 \lim _{n \rightarrow \infty} \frac{n(n+1)(2 n+1)}{6 n^3}=4 \lim _{n \rightarrow \infty} \frac{1 \cdot\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right)}{6} \\
& =4 \times \frac{2}{6}=\frac{4}{3}
\end{aligned}\)
Hence, option (b) is correct.
& \lim _{n \rightarrow \infty} \frac{2^2+4^2+6^2+\ldots+(2 n)^2}{n^3} \\
& =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.
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