A) 196
B) 197
C) 198
D) 199
Correct Answer: B
Solution :
Let \[{{(\sqrt{2}+1)}^{6}}=k+f,\]where k is integral part and f the fraction\[(0\le f<1)\]. Let\[{{(\sqrt{2}-1)}^{6}}=f',\,(0<f'<1)\], Since \[0<(\sqrt{2}-1)<1\] Now, \[k+f+{f}'={{(\sqrt{2}+1)}^{6}}+{{(\sqrt{2}-1)}^{6}}\] \[=2\left\{ {{\,}^{6}}{{C}_{0}}{{.2}^{3}}+{{\,}^{6}}{{C}_{2}}{{.2}^{2}}+{{\,}^{6}}{{C}_{4}}.2+{{\,}^{6}}{{C}_{6}} \right\}=198\] ?..(i) \[\therefore \,\,\,f+{f}'=198-k=\] an integer But, \[0\le f<1\]and \[0<{f}'<1\Rightarrow 0<(f+{f}')<2\] \[\Rightarrow f+{f}'=1\], \[(\because f+{f}'\]is an integer) \ By (i), I =198 ? (\[f+{f}'\]) = 198 ? 1 = 197.
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