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Question Number 9488 by Joel575 last updated on 10/Dec/16

lim_(x→∞) (((√(2016)) + 4(√(2016)) + 9(√(2016)) + ... + x^2 (√(2016)))/x^3 )

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{2016}}\:+\:\mathrm{4}\sqrt{\mathrm{2016}}\:+\:\mathrm{9}\sqrt{\mathrm{2016}}\:+\:...\:+\:{x}^{\mathrm{2}} \sqrt{\mathrm{2016}}}{{x}^{\mathrm{3}} } \\ $$

Answered by mrW last updated on 10/Dec/16

=lim_(x→∞) (((√(2016))×(1^2 + 2^2 +3^2 + ... + x^2 ))/x^3 ) =lim_(x→∞) ((√(2016))/x^3 )×((x(x+1)(2x+1))/6) =lim_(x→∞) ((√(2016))/6)×(1+(1/x))(2+(1/x)) =((√(2016))/3)=4(√(14))

$$=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{2016}}×\left(\mathrm{1}^{\mathrm{2}} \:+\:\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +\:...\:+\:{x}^{\mathrm{2}} \right)}{{x}^{\mathrm{3}} }\: \\ $$$$=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{2016}}}{\mathrm{x}^{\mathrm{3}} }×\frac{\mathrm{x}\left(\mathrm{x}+\mathrm{1}\right)\left(\mathrm{2x}+\mathrm{1}\right)}{\mathrm{6}} \\ $$$$=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{2016}}}{\mathrm{6}}×\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}}\right)\left(\mathrm{2}+\frac{\mathrm{1}}{\mathrm{x}}\right) \\ $$$$=\frac{\sqrt{\mathrm{2016}}}{\mathrm{3}}=\mathrm{4}\sqrt{\mathrm{14}} \\ $$

Commented by Joel575 last updated on 11/Dec/16

thank you very much

$${thank}\:{you}\:{very}\:{much} \\ $$

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