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 $(\sqrt{2}+1{)}^6$ = k + f, where k is integral part and f

the fraction (0 ≤ f < 1).

Let $(\sqrt{2}-1{)}^6$ = f, (0 < f < 1),

Since 0 < ($\sqrt{2}$ - 1) < 1

Now, k + f + f' = $(\sqrt{2}+1{)}^6$ + $(\sqrt{2}-1{)}^6$

= 2{${}^6{C}_0{.2}^3{+}^6{C}_2{.2}^2{+}^6{C}_4.2{+}^6{C}_6$} = 198 ..........(i)

∴ f + f' = 198 - k = an integer

But, 0 ≤ f < 1 and 0 < f' < 1 ⇒ 0 < (f + f') < 2

⇒ f + f' = 1, ( ∵ f + f' is an integer)

∴ By (i), l = 198 - (f + f') = 198 - 1 = 197.