Tính tổng $A=(2+\frac{1}{2})+(2^2+\frac{1}{2^2})^2+...+(2^{2018}+\frac{1}{2^{2018}})^{2018}$
$A=(2+\frac{1}{2})+(2^2+\frac{1}{2^2})^2+...+(2^{2018}+\frac{1}{2^{2018}})^{2018}$
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Started By trambau, 01-06-2018 - 21:02
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