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

Option: 1

196


Option: 2

197


Option: 3

198


Option: 4

199


Answers (1)

best_answer

Let (\sqrt{2}+1)^6=k+f where k is integral part and f the fraction (0 \leq f<1).

Let (\sqrt{2}-1)^6=f^{\prime},(0<f<1)

Since, 0<(\sqrt{2}-1)<1

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

=2\left\{{ }^6 C_0 \cdot 2^3+{ }^6 C_2 \cdot 2^2+{ }^6 C_4 \cdot 2+{ }^6 C_6\right\}=198      .....(i)

\therefore f+f^{\prime}=198-k= an integer

But  \quad 0 \leq f<1 and 0<f^{\prime}<1 \Rightarrow 0<\left(f+f^{\prime}\right)<2

\Rightarrow f+f^{\prime}=1, (f+f^{\prime} is an integer)

∴ By  (\mathrm{i}), I=198-\left(f^{f+f^{\prime}}\right)=198-1=197.

 

Posted by

Ritika Jonwal

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