Tính I = lim ((căn bậc hai của (2+4+...+2n)+n)/((căn bậc ba của (48(1^2+2^2+...+n^2))+2n)

* Hướng dẫn giải

Ta có $\mathrm{I}=\lim \frac{\sqrt{2+4+...+2\mathrm{n}}+\mathrm{n}}{\sqrt[3]{48\left(1^2+2^2+...+{\mathrm{n}}^2\right)}+2\mathrm{n}}$$=\lim \frac{\sqrt{2\left(1+2+...+\mathrm{n}\right)}+\mathrm{n}}{\sqrt[3]{48\left(1^2+2^2+...+{\mathrm{n}}^2\right)}+2\mathrm{n}}$

Mà $1+2+...+\mathrm{n}=\frac{\mathrm{n}(\mathrm{n}+1)}{2}$; $1^2+2^2+...+{\mathrm{n}}^2=\frac{\mathrm{n}(\mathrm{n}+1)(2\mathrm{n}+1)}{6}$

Suy ra $\mathrm{I}=\lim \frac{\sqrt{2\frac{\mathrm{n}(\mathrm{n}+1)}{2}}+\mathrm{n}}{\sqrt[3]{48\left(\frac{\mathrm{n}(\mathrm{n}+1)(2\mathrm{n}+1)}{6}\right)}+2\mathrm{n}}$$=\lim \frac{\sqrt{{\mathrm{n}}^2\left(1+\frac{1}{\mathrm{n}}\right)}+\mathrm{n}}{\sqrt[3]{8{\mathrm{n}}^3\left(1+\frac{1}{\mathrm{n}}\right)\left(2+\frac{1}{\mathrm{n}}\right)}+2\mathrm{n}}$

$=\lim \frac{\mathrm{n}\left(\sqrt{\left(1+\frac{1}{\mathrm{n}}\right)}+1\right)}{\mathrm{n}\left(\sqrt[3]{8\left(1+\frac{1}{\mathrm{n}}\right)\left(2+\frac{1}{\mathrm{n}}\right)}+2\right)}$$=\frac{\sqrt{1}+1}{\sqrt[3]{8}+2}=\frac{1}{2}$