# Expand (x + 2/x)^5 using binomial theorem?

${\left(\mathrm{x}+\frac{2}{\mathrm{x}}\right)}^{5}\phantom{\rule{0ex}{0ex}}={\mathrm{x}}^{5}+{}^{5}\mathrm{C}_{1}{\mathrm{x}}^{4}{\left(\frac{2}{\mathrm{x}}\right)}^{1}+{}^{5}\mathrm{C}_{2}{\mathrm{x}}^{3}{\left(\frac{2}{\mathrm{x}}\right)}^{2}+{}^{5}\mathrm{C}_{3}{\mathrm{x}}^{2}{\left(\frac{2}{\mathrm{x}}\right)}^{3}+{}^{5}\mathrm{C}_{4}{\mathrm{x}}^{1}{\left(\frac{2}{\mathrm{x}}\right)}^{4}+{\left(\frac{2}{\mathrm{x}}\right)}^{5}\phantom{\rule{0ex}{0ex}}={\mathrm{x}}^{5}+\frac{5!}{1!4!}{\mathrm{x}}^{4}\left(\frac{2}{\mathrm{x}}\right)+\frac{5!}{2!3!}{\mathrm{x}}^{3}\left(\frac{4}{{\mathrm{x}}^{2}}\right)+\frac{5!}{2!3!}{\mathrm{x}}^{2}\left(\frac{8}{{\mathrm{x}}^{3}}\right)+\frac{5!}{4!1!}{\mathrm{x}}^{1}\left(\frac{16}{{\mathrm{x}}^{4}}\right)+\frac{32}{{\mathrm{x}}^{5}}\phantom{\rule{0ex}{0ex}}={\mathbf{x}}^{\mathbf{5}}\mathbf{+}\mathbf{10}{\mathbf{x}}^{\mathbf{3}}\mathbf{+}\mathbf{40}\mathbf{x}\mathbf{+}\frac{\mathbf{80}}{\mathbf{x}}\mathbf{+}\frac{\mathbf{80}}{{\mathbf{x}}^{\mathbf{3}}}\mathbf{+}\frac{\mathbf{32}}{{\mathbf{x}}^{\mathbf{5}}}$

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go to hell i don't know
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