The sum of all rational terms in the expansion of (1+2^1/3+3^1/2)^6 is equal to _________.

Question

The sum of all rational terms in the expansion of ${(1+{2}^{1/3}+{3}^{1/2})}^{6}$ is equal to \_\_\_\_\_\_\_\_\_.

Answer: 612

Accepted Answer

${(1+{2}^{\frac{1}{3}}+{3}^{\frac{1}{2}})}^{6}$

$=\frac{|\, \underset{}{\, 6\, }}{|\, \underset{}{\, {r}_{1}\, }|\, \underset{}{\, {r}_{2}\, }|\, \underset{}{\, {r}_{3}\, }}(1{)}^{{r}_{1}}(2{)}^{\frac{{r}_{2}}{3}}(3{)}^{\frac{{r}_{3}}{2}}$

$\begin{gathered} {r}_{1} \quad  {r}_{2} \quad  {r}_{3} \ 6 \quad  0 \quad  0 \ 4 \quad  0 \quad  2 \ 2 \quad  0 \quad  4 \ 0 \quad  0 \quad  6 \ 3 \quad  3 \quad  0 \ 1 \quad  3 \quad  2 \ 0 \quad  6 \quad  0 \end{gathered}$

$=\frac{|\, \underset{}{\, 6\, }}{|\, \underset{}{\, 6\, }|\, \underset{}{\, 0\, }|\, \underset{}{\, 0\, }}+\frac{|\, \underset{}{\, 6\, }}{|\, \underset{}{\, 4\, }|\, \underset{}{\, 0\, }|\, \underset{}{\, 2\, }}(3)+\frac{|\, \underset{}{\, 6\, }}{|\, \underset{}{\, 2\, }|\, \underset{}{\, 0\, }|\, \underset{}{\, 4\, }}(3{)}^{2}+\frac{|\, \underset{}{\, 6\, }}{|\, \underset{}{\, 0\, }|\, \underset{}{\, 0\, }|\, \underset{}{\, 6\, }}(3{)}^{3}+\frac{|\, \underset{}{\, 6\, }}{|\, \underset{}{\, 3\, }|\, \underset{}{\, 3\, }|\, \underset{}{\, 0\, }}(2)+\frac{|\, \underset{}{\, 6\, }}{|\, \underset{}{\, 1\, }|\, \underset{}{\, 3\, }|\, \underset{}{\, 2\, }}(2{)}^{1}(3{)}^{1}+\frac{|\, \underset{}{\, 6\, }}{|\, \underset{}{\, 0\, }|\, \underset{}{\, 6\, }|\, \underset{}{\, 0\, }}(2{)}^{2}$

$=1+45+135+27+40+360+4=612$

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