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If $n$ is a positive integer, then $(\sqrt{3}+1)^{2 n}-(\sqrt{3}-1)^{2 n}$ is
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Verified Answer
The correct answer is:
an irrational number
an irrational number
$\begin{aligned}
& (\sqrt{3}+1)^{2 n}-(\sqrt{3}-1)^{2 n}=\left[(\sqrt{3}+1)^2\right]^n-\left[(\sqrt{3}-1)^2\right]^n=(4+2 \sqrt{3})^n-(4-2 \sqrt{3})^n \\
& =2^n\left[(2+\sqrt{3})^n-(2-\sqrt{3})^n\right] \\
& =2^n\left\{\left[{ }^n C_0 2^n+{ }^n C_1 2^{n-1} \sqrt{3}+{ }^n C_2 2^{n-2} 3+\cdots \cdot\right]-\left[{ }^n C_0 2^n-{ }^n C_1 2^{n-1} \sqrt{3}+{ }^n C_2 2^{n-2} 3-\cdots \cdot\right]\right\} \\
& =2^{n+1}\left[{ }^n C_1 2^{n-1} \sqrt{3}+{ }^n C_3 2^{n-3} 3 \sqrt{3}+\cdots \cdot\right]=2^{n+1} \sqrt{3} \text { (some integer) }
\end{aligned}$
Which is irrational
& (\sqrt{3}+1)^{2 n}-(\sqrt{3}-1)^{2 n}=\left[(\sqrt{3}+1)^2\right]^n-\left[(\sqrt{3}-1)^2\right]^n=(4+2 \sqrt{3})^n-(4-2 \sqrt{3})^n \\
& =2^n\left[(2+\sqrt{3})^n-(2-\sqrt{3})^n\right] \\
& =2^n\left\{\left[{ }^n C_0 2^n+{ }^n C_1 2^{n-1} \sqrt{3}+{ }^n C_2 2^{n-2} 3+\cdots \cdot\right]-\left[{ }^n C_0 2^n-{ }^n C_1 2^{n-1} \sqrt{3}+{ }^n C_2 2^{n-2} 3-\cdots \cdot\right]\right\} \\
& =2^{n+1}\left[{ }^n C_1 2^{n-1} \sqrt{3}+{ }^n C_3 2^{n-3} 3 \sqrt{3}+\cdots \cdot\right]=2^{n+1} \sqrt{3} \text { (some integer) }
\end{aligned}$
Which is irrational
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