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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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} = $

Similar Questions

Let $[x]$ denote the greatest integer less than or equal to $x$. Then,evaluate the limit: $\mathop {\lim }\limits_{x \to 0} \,\frac{{\tan \,(\pi \,{{\sin }^2}\,x) + \,{{(\left| x \right|\, - \,\sin \,(x\,[x]))}^2}}}{{{x^2}}}$

Evaluate the given limit: $\mathop {\lim }\limits_{z \to 1} \frac{z^{1/3}-1}{z^{1/6}-1}$

$\mathop {Lim}\limits_{n \to \infty } \frac{{{1^2}n + {2^2}(n - 1) + {3^2}(n - 2) + \dots + {n^2} \cdot 1}}{{{1^3} + {2^3} + {3^3} + \dots + {n^3}}}$ is equal to :

$\lim _{x \rightarrow 0} \frac{\sin |x|}{x}$ is equal to

$\mathop {\lim }\limits_{x \to \infty } (\sqrt {{x^2} + 1} - x)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module