Binomial theorem for positive integral Index

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Question

The number of rational terms in the expansion of $\left(1+\sqrt{2}+\sqrt[3]{5}{\right)}^{6}$ is

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Solution

A term in the expansion of $\left(1+\sqrt{2}+\sqrt[3]{5}{\right)}^{6}$ isof the form$\frac{6!}{r!s!\left(6-r-s\right)!}\left(1{\right)}^{6-r-s}\left(\sqrt{2}{\right)}^{r}\left(\sqrt[3]{5}{\right)}^{s}$$=\frac{6!}{r!s!\left(6-r-s\right)!}\left({2}^{r/2}\right)\left({5}^{s/3}\right)$This term will be rational if ${2}^{r/2}$ and  are both rational numbers. This is possible if and only if  $r$ is $a$ multiple of $2$ and s is a multiple of $3.$ Possible values of r are  and 6 whereas the possible values of $s$ are   and  Also note that $0\le r+s\le 6$For $r=0,s$ can take value  or $6$For can take value 0 or $3$For can take value $0$For  can take value $0.$Thus, the number of rational terms is  $3+2+1+1=7$

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The coefficient of x4 in the expansion of ${\left(1+\mathrm{x}+{\mathrm{x}}^{2}+{\mathrm{x}}^{3}\right)}^{\mathrm{n}}$ is