Question

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

• A. 196
• B. 197
• C. 198
• D. 199

Answer 1

The correct option is B 197

Let ${(\sqrt{2}+1)}^6=I+F,$ Where I is an integer and 0< F<1.
$f=\sqrt{2}-1=\frac{1}{\sqrt{2}+1}\Rightarrow 0<\sqrt{2}-1<1\Rightarrow 0<f<1$
Also $I+F+f={(\sqrt{2}+1)}^6+{(\sqrt{2}-1)}^6$
= $2{[}^6{C}_0{.2}^3{+}^6{C}_2{.2}^2{+}^6{C}_4.2{+}^6{C}_6]$
$=2(8+60+30+1)=198$
Hence, $F+f=198-I$ is an integer. But $0<F+f<2$.
$\therefore F+f=1$, and thus, $I=197$