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(A) The general term in the expansion of $(1 + x^{\log_2 x})^5$ is given by $T_{r+1} = ^5C_r (x^{\log_2 x})^r$.For the third term,$r = 2$,so $T_3 = ^5

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$1/4$

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(A) The general term in the expansion of $(1 + x^{\log_2 x})^5$ is given by $T_{r+1} = ^5C_r (x^{\log_2 x})^r$.For the third term,$r = 2$,so $T_3 = ^5C_2 (x^{\log_2 x})^2$.Given $T_3 = 2560$,we have $10 (x^{\log_2 x})^2 = 2560$.$(x^{\log_2 x})^2 = 256$.Taking the logarithm to the base $2$ on both sides:$2 \log_2 (x^{\log_2 x}) = \log_2 (256)$.$2 (\log_2 x)(\log_2 x) = 8$.$(\log_2 x)^2 = 4$.$\log_2 x = \pm 2$.If $\log_2 x = 2$,then $x = 2^2 = 4$.If $\log_2 x = -2$,then $x = 2^{-2} = 1/4$.Thus,a possible value of $x$ is $1/4$.

If the third term in the binomial expansion of $(1 + x^{\log_2 x})^5$ equals $2560$,then a possible value of $x$ is

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