Question

The integral part of ${(\sqrt{2}+1)}^6$ is

• A. 197
• B. 196
• C. 175
• D. 176

Answer 1

The correct option is A 197

$Let(\sqrt{2}+1{)}^6=1+F,whereIistheintegralpart.Then0<F<1Letf=(\sqrt{2}+1{)}^{6}0<\sqrt{2}-1<1\Rightarrow 0<(\sqrt{2}+1{)}^6<1\Rightarrow 0<f<10<F<1,0<f<1\Rightarrow 0<F+f<2I+F+f=(\sqrt{2}+1{)}^6+(\sqrt{2}-1{)}^6=2{[}^6{C}_0(\sqrt{2}{)}^6{+}^6{C}_2\left(\sqrt{2}{)}^4{+}^6{C}_4\right(\sqrt{2}{)}^2{+}^6{C}_6]=2[8+15\left(4\right)+15\left(2\right)+1]=2\times 99=198I+F+f,Iareintegers\Rightarrow F+fisaninteger0<F+f<2,F+fisaninteger\Rightarrow F+f=1I+F+f=198\Rightarrow I+1=198\Rightarrow I=197$