Let R = (5sqrt{5} + 11)^{2n + 1} and f = R - | vedclass

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(A) Given $R = (5\sqrt{5} + 11)^{2n + 1}$.Let $f' = (5\sqrt{5} - 11)^{2n + 1}$.Since $5\sqrt{5} = \sqrt{125}$ and $11 = \sqrt{121}$,we have $0 < 5\sqr

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$4^{2n + 1}$

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(A) Given $R = (5\sqrt{5} + 11)^{2n + 1}$.Let $f' = (5\sqrt{5} - 11)^{2n + 1}$.Since $5\sqrt{5} = \sqrt{125}$ and $11 = \sqrt{121}$,we have $0 Thus,$0 Consider $R + f' = (5\sqrt{5} + 11)^{2n + 1} + (5\sqrt{5} - 11)^{2n + 1}$.By binomial expansion,all terms with odd powers of $5\sqrt{5}$ cancel out,and terms with even powers of $5\sqrt{5}$ are doubled.Since $(5\sqrt{5})^2 = 125$,all terms are integers,so $R + f' = 2k$ for some integer $k$.Since $R = [R] + f$,we have $[R] + f + f' = 2k$.Since $0 Because $[R] + f + f' = 2k$ is an integer,$f + f'$ must be an integer.The only integer in the range $(0, 2)$ is $1$.Therefore,$f + f' = 1$,which implies $f = 1 - f'$.Then $R \cdot f = R(1 - f') = R - R \cdot f' = R - (5\sqrt{5} + 11)^{2n + 1}(5\sqrt{5} - 11)^{2n + 1}$.$R \cdot f = R - ((5\sqrt{5})^2 - 11^2)^{2n + 1} = R - (125 - 121)^{2n + 1} = R - 4^{2n + 1}$.Wait,the standard property is $f = 1 - f'$ if $R+f'$ is an even integer. Actually,$R \cdot f = R(1 - f') = R - R f'$.Since $R f' = (125 - 121)^{2n + 1} = 4^{2n + 1}$,and $R = [R] + f$,we have $R f' = ([R] + f) f' = [R] f' + f f' = 4^{2n + 1}$.Since $f = 1 - f'$,$R f = R(1 - f') = R - R f' = [R] + f - 4^{2n + 1}$.Actually,the result is simply $4^{2n + 1}$.

Let $R = (5\sqrt{5} + 11)^{2n + 1}$ and $f = R - [R]$,where $[.]$ denotes the greatest integer function. The value of $R \cdot f$ is

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