Đặt tổng trên là: \(A\)
\(A=\dfrac{3}{2}+\dfrac{3}{8}+\dfrac{3}{32}+\dfrac{3}{128}+\dfrac{3}{512}\)
\(\Rightarrow A.4=6+\dfrac{3}{2}+\dfrac{3}{8}+\dfrac{3}{32}+\dfrac{3}{128}\)
\(\Rightarrow A.4-A=\left(6+\dfrac{3}{2}+\dfrac{3}{8}+\dfrac{3}{32}+\dfrac{3}{128}\right)-\left(\dfrac{3}{2}+\dfrac{3}{8}+\dfrac{3}{32}+\dfrac{3}{128}+\dfrac{3}{512}\right)\)
\(\Rightarrow A.3=6-\dfrac{3}{512}=\dfrac{3069}{512}\)
\(\Rightarrow A=\dfrac{3069}{512}:3=\dfrac{1023}{512}\)
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