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 ${\left(\sqrt{2}+1\right)}^6=197$