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

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Question

(B) ધારો કે પદાવલિ $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)\righ

Vedclass · Accepted Answer

$\frac{144}{5}$

Vedclass · Answer

(B) ધારો કે પદાવલિ $L = \lim _{n ightarrow \infty} \frac{\left[6^2(1^2+2^2+\ldots+n^2)ight]^2}{[5(1+2+\ldots+n)]\left[2^3(1^3+2^3+\ldots+n^3)ight]}$ છે.સરવાળાના સૂત્રો $\sum_{k=1}^n k = \frac{n(n+1)}{2}$,$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$,અને $\sum_{k=1}^n k^3 = \frac{n^2(n+1)^2}{4}$ નો ઉપયોગ કરતા:અંશ $= 36^2 \left[\frac{n(n+1)(2n+1)}{6}ight]^2 = 1296 \frac{n^2(n+1)^2(2n+1)^2}{36} = 36 n^2(n+1)^2(2n+1)^2$.છેદ $= 5 \left[\frac{n(n+1)}{2}ight] 	imes 8 \left[\frac{n^2(n+1)^2}{4}ight] = 5 	imes \frac{n(n+1)}{2} 	imes 2n^2(n+1)^2 = 5 n^3(n+1)^3$.$L = \lim _{n ightarrow \infty} \frac{36 n^2(n+1)^2(2n+1)^2}{5 n^3(n+1)^3} = \lim _{n ightarrow \infty} \frac{36 n^2(n+1)^2(4n^2)}{5 n^3(n+1)^3} = \lim _{n ightarrow \infty} \frac{144 n^6}{5 n^6} = \frac{144}{5}$.

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

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