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The number of rational terms in the binomial expansion of $(\sqrt[4]{5}+\sqrt[5]{4})^{100}$ is
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Verified Answer
The correct answer is:
6
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.
$$
\begin{aligned}
(\sqrt[4]{5}+\sqrt[5]{4})^{100} & =\sum_{r=0}^{100}{ }^{100} C_r(\sqrt[4]{5})^{100-r}(\sqrt[5]{4})^r \\
& =\sum_{r=0}^{100}{ }^{100} C_r 5 \frac{100-r}{4} 4^{\frac{r}{5}}=\sum_{r=0}^{100} T_{r+1} \\
T_{r+1} & ={ }^{100} C_r 5 \frac{100-r}{4} 4^{\frac{r}{5}}
\end{aligned}
$$
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.
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