MCQ
The greatest integer less than or equal to ${(\sqrt 2 + 1)^6}$ is
- A$196$
- ✓$197$
- C$198$
- D$199$
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\left\{ {{\,^6}{C_0}{{.2}^3} + {\,^6}{C_2}{{.2}^2} + {\,^6}{C_4}.2 + {\,^6}{C_6}} \right\} = 198$…..(i)
$\therefore \,\,\,f + f' = 198 - k = $ an integer
But, $0 \le f < 1$ and $0 < f' < 1 \Rightarrow 0 < (f + f') < 2$
$ \Rightarrow f + f' = 1$, is an integer
By (i), $I =198 -(f + f') = 198 -1 = 197$.
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