The greatest integer less than or equal to (s | vedclass

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(B) Let $(\sqrt{2} + 1)^6 = I + f$,where $I$ is the integral part and $f$ is the fractional part $(0 \le f < 1)$.Let $f' = (\sqrt{2} - 1)^6$. Since $0

Vedclass · Accepted Answer

$197$

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(B) Let $(\sqrt{2} + 1)^6 = I + f$,where $I$ is the integral part and $f$ is the fractional part $(0 \le f Let $f' = (\sqrt{2} - 1)^6$. Since $0 Consider the sum $S = (\sqrt{2} + 1)^6 + (\sqrt{2} - 1)^6$.Using the binomial expansion,$S = 2 \left[ \binom{6}{0}(\sqrt{2})^6 + \binom{6}{2}(\sqrt{2})^4 + \binom{6}{4}(\sqrt{2})^2 + \binom{6}{6} \right]$.$S = 2 [1 \cdot 8 + 15 \cdot 4 + 15 \cdot 2 + 1 \cdot 1] = 2 [8 + 60 + 30 + 1] = 2 [99] = 198$.Thus,$I + f + f' = 198$.Since $0 Therefore,$I = 198 - 1 = 197$.

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

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