lim _{n rightarrow infty} frac{2^2+4^2+6^2+ld | vedclass

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Question

(B) The given expression is $\lim _{n \rightarrow \infty} \frac{2^2+4^2+6^2+\ldots+(2 n)^2}{n^3}$.We can factor out $2^2 = 4$ from the numerator:$= \l

Vedclass · Accepted Answer

$\frac{4}{3}$

Vedclass · Answer

(B) The given expression is $\lim _{n ightarrow \infty} \frac{2^2+4^2+6^2+\ldots+(2 n)^2}{n^3}$.We can factor out $2^2 = 4$ from the numerator:$= \lim _{n ightarrow \infty} \frac{4(1^2+2^2+3^2+\ldots+n^2)}{n^3}$.Using the formula for the sum of squares of the first $n$ natural numbers,$\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$:$= 4 \lim _{n ightarrow \infty} \frac{n(n+1)(2n+1)}{6n^3}$.$= \frac{4}{6} \lim _{n ightarrow \infty} \frac{n \cdot n(1+\frac{1}{n}) \cdot n(2+\frac{1}{n})}{n^3}$.$= \frac{2}{3} \lim _{n ightarrow \infty} (1+\frac{1}{n})(2+\frac{1}{n})$.As $n ightarrow \infty$,$\frac{1}{n} ightarrow 0$,so the limit is $\frac{2}{3} 	imes (1)(2) = \frac{4}{3}$.Thus,the correct option is $B$.

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

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