Question

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

• A. $195$
• B. $196$
• C. $197$
• D. $198$

Answer 1

The correct option is C $197$

${(\sqrt{2}+1)}^6=A$

$\left[A\right]$ is the greatest integer less than $A$.

Let $f$ be some unknown fraction.

Hence, we can write $\left[A\right]+f=A$.

Now, $(\sqrt{2}-1)<1{(\sqrt{2}-1)}^6<1$

Let, ${(\sqrt{2}-1)}^6$ be denoted by $f_1$.

So, ${(\sqrt{2}+1)}^6+{(\sqrt{2}-1)}^6=\left[A\right]+f+f_1\Rightarrow 2\left({}^6{C}_0.{\left(\sqrt{2}\right)}^6{+}^6{C}_2.{\left(\sqrt{2}\right)}^4{+}^6{C}_6\right)=\left[A\right]+f+f_1\Rightarrow 2(8+60+31)=\left[A\right]+f+f_1\Rightarrow 198=\left[A\right]+f+f_1$

Now, $0<f<10<f_1<10<f+f_1<2$

But as $198-\left[A\right]=f+f_1$, So, $f+f_1$ has to be an integer.

As, the only integer in $(0,2)$ is 1, hence $f+f_1=1$

So, $\left[A\right]=197$