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  4. The number of integral terms in the expansion of (5(1/6) + 7(1/9))1824 is

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Q. The number of integral terms in the expansion of $\left(5^{\frac{1}{6}} + 7^{\frac{1}{9}}\right)^{1824}$ is

NTA AbhyasNTA Abhyas 2020Binomial Theorem

Solution:

$T_{r + 1}=^{1824}C_{r}\left(5^{\frac{1}{6}}\right)^{1824 - r}\left(7^{\frac{1}{9}}\right)^{r}$
$=^{1824}C_{r}\left(5^{\frac{1824 - r}{6}}\right)\left(7^{\frac{r}{9}}\right)$
$=^{1824}C_{r}\left(5^{304 - \frac{r}{6}}\right)\left(7^{\frac{r}{9}}\right)$
Both $\frac{r}{6}$ and $\frac{r}{9}$ are integers, if $r$ is a multiple of $18$
$\Rightarrow r=0,18,36,......,1818$
Let, the number of terms be $n$
$\Rightarrow 1818=0+\left(n - 1\right)18\Rightarrow n=102.$