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The number of integral terms in the expansion of $(\sqrt{3}+\sqrt[8]{5})^{256}$ is
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Verified Answer
The correct answer is:
$33$
$$
\begin{aligned}
& (\sqrt{3}+\sqrt[8]{5})^{256} \\
T_{r+1} & ={ }^{256} C_r\left(3^{1 / 2}\right)^{256-r}\left(5^{1 / 8}\right)^r
\end{aligned}
$$
For integral terms, $r$ must be divisible by 2 and 8 .
$\therefore \quad r$ must be divisible by 8
i.e. $r=0,8,16, \ldots . ., 256$
$$
\begin{aligned}
& t_n=a+(n-1) d \\
& \Rightarrow 256=0+(n-1) 8 \\
& \Rightarrow 32=n-1 \\
& \therefore n=33 .
\end{aligned}
$$
\begin{aligned}
& (\sqrt{3}+\sqrt[8]{5})^{256} \\
T_{r+1} & ={ }^{256} C_r\left(3^{1 / 2}\right)^{256-r}\left(5^{1 / 8}\right)^r
\end{aligned}
$$
For integral terms, $r$ must be divisible by 2 and 8 .
$\therefore \quad r$ must be divisible by 8
i.e. $r=0,8,16, \ldots . ., 256$
$$
\begin{aligned}
& t_n=a+(n-1) d \\
& \Rightarrow 256=0+(n-1) 8 \\
& \Rightarrow 32=n-1 \\
& \therefore n=33 .
\end{aligned}
$$
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