The sum of all coefficients in the binomial expansion of ${(1 + 2 x)}^{k}$ is 6561. Let R = ${(1 + 2 x)}^{k}$ $= I + F$ Where $I \in N$ and $0 < F < 1 .$ If $x = \frac{1}{\sqrt{2}},$ then $1 - \frac{F}{1 + {(\sqrt{2} - 1)}^{4}} =$

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$(3 \sqrt{2} - 4)$

$4 (3 \sqrt{2} + 4)$

${(\sqrt{2} - 1)}^{4}$

answer is C.

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Detailed Solution

$3^{k} = 6561 \Rightarrow k = 8 R = {(1 + 2 x)}^{8} = I + F, x = \frac{1}{\sqrt{2}} R = {(1 + \sqrt{2})}^{8} = I + F Let f = {(1 - \sqrt{2})}^{8} I + F + f = 2 ({}^{8}{C_{0} + {}^{8}{C_{2}} {(\sqrt{2})}^{6} + . . . . . .}) = Integer$

$(F + f)$ is an Integer

$0 < F + f < 2 F + f = 1 \Rightarrow f = 1 - F$

$\begin{array}{rcl} 1 - \frac{F}{1 + {(\sqrt{2} - 1)}^{4}} & = & \frac{1 + {(\sqrt{2} - 1)}^{4} - F}{1 + {(\sqrt{2} - 1)}^{4}} \\ = & \frac{f + {(\sqrt{2} - 1)}^{4}}{1 + {(\sqrt{2} - 1)}^{4}} \\ = & \frac{{(\sqrt{2} - 1)}^{8} + {(\sqrt{2} - 1)}^{4}}{1 + {(\sqrt{2} - 1)}^{4}} \\ = & {(\sqrt{2} - 1)}^{4} \end{array}$