The number of rational terms in the binomial expansion of $(\sqrt[4]{5}+\sqrt[5]{4})^{100}$ is
The number of rational terms in the binomial expansion of $(\sqrt[4]{5}+\sqrt[5]{4})^{100}$ is
50
5
6
51
Solution
We have,
$
\begin{aligned}
(\sqrt[4]{5}+\sqrt[5]{4})^{100} & =\sum_{r=0}^{100}{ }^{100} C_r(\sqrt[4]{5})^{100-r}(\sqrt[5]{4})^r \\
& =\sum_{r=0}^{100}{ }^{100} C_r 5 \frac{100-r}{4} 4^{\frac{r}{5}}=\sum_{r=0}^{100} T_{r+1} \\
T_{r+1} & ={ }^{100} C_r 5 \frac{100-r}{4} 4^{\frac{r}{5}}
\end{aligned}
$
Where, $\quad T_{r+1}={ }^{100} C_r 5 \frac{100-r}{4} 4^{\frac{r}{5}}$
Clearly, $T_{r+1}$ will be an integer if $\frac{100-r}{4}$ and $\frac{r}{5}$ are integers. This is possible when
$100-r$ is a multiple of 4 and $r$ is a multiple of 5 $\Rightarrow 100-r=0,4,8,12, \ldots, 96,100$ and
$
r=0,5,10, \ldots, 100
$
$\Rightarrow r=0,4,8,12, \ldots, 100$ and $r=0,5,10,100$ $\Rightarrow r=0,20,40,60,80,100$
Hence, there are 6 rational terms