lim _{n rightarrow infty} frac{left[6^2+12^2+ | vedclass

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

Question

(B) माना व्यंजक $L = \lim _{n \rightarrow \infty} \frac{\left[6^2(1^2+2^2+\ldots+n^2)\right]^2}{[5(1+2+\ldots+n)]\left[2^3(1^3+2^3+\ldots+n^3)\right]}

Vedclass · Accepted Answer

$\frac{144}{5}$

Vedclass · Answer

(B) माना व्यंजक $L = \lim _{n ightarrow \infty} \frac{\left[6^2(1^2+2^2+\ldots+n^2)ight]^2}{[5(1+2+\ldots+n)]\left[2^3(1^3+2^3+\ldots+n^3)ight]}$ है।योगफल सूत्रों $\sum_{k=1}^n k = \frac{n(n+1)}{2}$,$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$,और $\sum_{k=1}^n k^3 = \frac{n^2(n+1)^2}{4}$ का उपयोग करते हुए:अंश $= 36^2 \left[\frac{n(n+1)(2n+1)}{6}ight]^2 = 1296 \frac{n^2(n+1)^2(2n+1)^2}{36} = 36 n^2(n+1)^2(2n+1)^2$.हर $= 5 \left[\frac{n(n+1)}{2}ight] 	imes 8 \left[\frac{n^2(n+1)^2}{4}ight] = 5 	imes \frac{n(n+1)}{2} 	imes 2n^2(n+1)^2 = 5 n^3(n+1)^3$.$L = \lim _{n ightarrow \infty} \frac{36 n^2(n+1)^2(2n+1)^2}{5 n^3(n+1)^3} = \lim _{n ightarrow \infty} \frac{36 n^2(n+1)^2(4n^2)}{5 n^3(n+1)^3} = \lim _{n ightarrow \infty} \frac{144 n^6}{5 n^6} = \frac{144}{5}$.

$\lim _{n \rightarrow \infty} \frac{\left[6^2+12^2+18^2+\ldots+(6 n)^2\right]^2}{[5+10+15+\ldots+5 n]\left[2^3+4^3+6^3+\ldots+(2 n)^3\right]} =$

Similar Questions

यदि $\Delta(x) = \begin{vmatrix} e^x & -1 \\ \sin x - 1 & 1 \end{vmatrix}$ है,तो $\lim_{x \rightarrow 0} \frac{\Delta(x)}{x} = $

$\lim _{x \rightarrow 0} \frac{3^{\sin x}-2^{\tan x}}{\sin x}=$

$\mathop {Lim}\limits_{x \to 0} \frac{{\log _{{{\sin }^2}x}}\cos x}{{\log _{{{\sin }^2}\frac{x}{2}}}\cos \frac{x}{2}}$ का मान किसके बराबर है?

$\mathop {\lim }\limits_{x \to 3} \frac{{\sqrt {3x} - 3}}{{\sqrt {2x - 4} - \sqrt 2 }}$ का मान ज्ञात कीजिए।

Vedclass Test Series

Exam Paper Generator

Online Exam Module