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The greatest integer less than or equal to $(\sqrt{3}+2)^5$ is
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1594 Upvotes
Verified Answer
The correct answer is:
723
Let $(2+\sqrt{3})^5=Y+F$
$[\because Y$ is integer, $F$ is fraction]
$$
\because \quad 2-\sqrt{3} < 1 \Rightarrow(2-\sqrt{3})^5 < 1
$$
Let $\quad(2-\sqrt{3})^5=F^{\prime}$
$$
\begin{aligned}
\therefore \quad & Y+F+F^{\prime}=(2+\sqrt{3})^5+(2-\sqrt{3})^5 \\
& =2\left(2^5+{ }^5 C_2 2^3(3)+{ }^5 C_4 2 \cdot 3^2\right) \\
& Y+F+F^{\prime}=2(32+240+90) \\
& Y+F+F^{\prime}=724 \\
& Y=724-\left(F+F^{\prime}\right) \\
& Y=724-1=723 \quad\left[\because F+F^{\prime}=1\right]
\end{aligned}
$$
$[\because Y$ is integer, $F$ is fraction]
$$
\because \quad 2-\sqrt{3} < 1 \Rightarrow(2-\sqrt{3})^5 < 1
$$
Let $\quad(2-\sqrt{3})^5=F^{\prime}$
$$
\begin{aligned}
\therefore \quad & Y+F+F^{\prime}=(2+\sqrt{3})^5+(2-\sqrt{3})^5 \\
& =2\left(2^5+{ }^5 C_2 2^3(3)+{ }^5 C_4 2 \cdot 3^2\right) \\
& Y+F+F^{\prime}=2(32+240+90) \\
& Y+F+F^{\prime}=724 \\
& Y=724-\left(F+F^{\prime}\right) \\
& Y=724-1=723 \quad\left[\because F+F^{\prime}=1\right]
\end{aligned}
$$
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