According to me the integral part of $(\sqrt 6 +2)^n$ is an even number. Here $n$ is a natural number.
But in my book it is written that it not always correct.
How is this possible?
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It is not always correct because for $n=2$, we have $$(\sqrt{6}+2)^2 = 2^2+ 4 \sqrt{6} + 6 = 10 + 4 \sqrt{6} = 10 + \sqrt{96}$$ Because $81 < 96 < 100$, we have $9 < \sqrt{96} < 10$, hence $19< (\sqrt{6}+2)^2 < 20$ and hence the integral part of $(\sqrt{6}+2)^2$ is $19$, which is odd. Therefore, it is not true for all $n$.
wythagoras
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