Question Number 9488 by Joel575 last updated on 10/Dec/16
$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{2016}}\:+\:\mathrm{4}\sqrt{\mathrm{2016}}\:+\:\mathrm{9}\sqrt{\mathrm{2016}}\:+\:…\:+\:{x}^{\mathrm{2}} \sqrt{\mathrm{2016}}}{{x}^{\mathrm{3}} } \\ $$
Answered by mrW last updated on 10/Dec/16
$$=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{2016}}×\left(\mathrm{1}^{\mathrm{2}} \:+\:\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +\:…\:+\:{x}^{\mathrm{2}} \right)}{{x}^{\mathrm{3}} }\: \\ $$$$=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{2016}}}{\mathrm{x}^{\mathrm{3}} }×\frac{\mathrm{x}\left(\mathrm{x}+\mathrm{1}\right)\left(\mathrm{2x}+\mathrm{1}\right)}{\mathrm{6}} \\ $$$$=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{2016}}}{\mathrm{6}}×\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}}\right)\left(\mathrm{2}+\frac{\mathrm{1}}{\mathrm{x}}\right) \\ $$$$=\frac{\sqrt{\mathrm{2016}}}{\mathrm{3}}=\mathrm{4}\sqrt{\mathrm{14}} \\ $$
Commented by Joel575 last updated on 11/Dec/16
$${thank}\:{you}\:{very}\:{much} \\ $$