AP EAMCET · Maths · Sequences and Series
\(\lim _{n \rightarrow \infty} \frac{2^2+4^2+6^2+\ldots+(2 n)^2}{n^3}=\)
- A \(\frac{2}{3}\)
- B \(\frac{4}{3}\)
- C \(\frac{3}{2}\)
- D \(\frac{8}{7}\)
Answer & Solution
Correct Answer
(B) \(\frac{4}{3}\)
Step-by-step Solution
Detailed explanation
\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…
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