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The sum of rational terms in the expansion of $(2^{\large\frac{1}{5}}$$+\sqrt 3)^{20}$ is

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General term $T_{r+1}$ in the expansion of $(2^{\large\frac{1}{5}}$$+\sqrt 3)^{20}$ is
$^{20}C_r.(2^{\large\frac{1}{5}}$$)^{20-r}.\sqrt 3^r$
$=^{20}C_r.2^{4-\large\frac{r}{5}}$$.3^{\large\frac{r}{5}}$
Since $2$ and $3$ are relatively prime,
$2^{4-\large\frac{r}{5}}$$.3^{\large\frac{r}{5}}$ will be rational if both
$4-\large\frac{r}{5}$ and $\large\frac{r}{2}$ are integers.
$\Rightarrow\:\large\frac{r}{2},\frac{r}{5}$ are integers.
i.e., $r=0,10,20$
$\therefore\: $ sum of rational terms =
$^{20}C_0.2^4.3^0+^{20}C_{10}.2^2.3^5+^{20}C_{20}2^0.3^{10}$
which is not $71,85 or 97$
None of the above is the answer.
answered Oct 19, 2013 by rvidyagovindarajan_1
 

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