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The number of irrational terms in the binomial expansion of $\left(3^{1 / 5}+7^{1 / 3}\right)^{100}$ is
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The correct answer is:
94
General term of $\left(3^{1 / 5}+7^{1 / 3}\right)^{100}$ is given by $T_{\gamma+1}={ }^{100} C_{r}\left(3^{1 / 5}\right)^{100-r}\left(7^{1 / 3}\right)^{r}$
$={ }^{100} C_{1} \cdot 3^{\frac{100-r}{5}} \cdot 7^{\frac{r}{3}}$
For a rational term. $\frac{100-r}{5}$ and $\frac{r}{3}$ must be integer.
Hence, $r=0,15,30,45,60,75,90$
$\therefore$ There are seven rational terms. Hence, number of irrational terms
$$
=101-7=94
$$
$={ }^{100} C_{1} \cdot 3^{\frac{100-r}{5}} \cdot 7^{\frac{r}{3}}$
For a rational term. $\frac{100-r}{5}$ and $\frac{r}{3}$ must be integer.
Hence, $r=0,15,30,45,60,75,90$
$\therefore$ There are seven rational terms. Hence, number of irrational terms
$$
=101-7=94
$$
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