Question

The greatest integer less than $\left(\sqrt{2} + 1\right)^{6}$ is equal to

• A. $196$
• B. $197$
• C. $198$
• D. $199$

Answer 1

Answer: (B)

$\left(\sqrt{2} + 1\right)^{6}+\left(\sqrt{2} - 1\right)^{6}=2\left(^{6} C_{0} 2^{3} + ^{6} C_{2} 2^{2} + ^{6} C_{4} 2^{1} + ^{6} C_{6} 2^{0}\right)$
$\left(\sqrt{2} + 1\right)^{6}+\left(\sqrt{2} - 1\right)^{6}=2\left(8 + 60 + 30 + 1\right)=198$
$\left(\sqrt{2} + 1\right)^{6}=198-\left(\sqrt{2} - 1\right)^{6}$
as $\sqrt{2}-1\approx 0.414\Rightarrow \left(\sqrt{2} - 1\right)^{6}\in \left(0,1\right)$
$\Rightarrow \left(\sqrt{2} + 1\right)^{6}=198-f,f\in \left(0,1\right)$
$\Rightarrow $ integral part of $\left(\sqrt{2} + 1\right)^{6}=197$