Đáp án:
Giải thích các bước giải:
$A=1+\dfrac{1}{2}+\dfrac{1}{2^{2}}+...+\dfrac{1}{2^{2019}}$
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$⇒\dfrac{1}{2}.A=\dfrac{1}{2}+\dfrac{1}{2^{2}}+\dfrac{1}{2^{3}}+...+\dfrac{1}{2^{2020}}$
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$⇒\dfrac{1}{2}.A-A=(\dfrac{1}{2}+\dfrac{1}{2^{2}}+\dfrac{1}{2^{3}}+...+\dfrac{1}{2^{2020}})-(1+\dfrac{1}{2}+\dfrac{1}{2^{2}}+...+\dfrac{1}{2^{2019}})$
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$⇒\dfrac{-1}{2}A=\dfrac{1}{2^{2020}}-1$
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$⇒\dfrac{-1}{2}.A=\dfrac{1-2^{2020}}{2^{2020}}$
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$⇒A=\dfrac{-2.(1-2^{2020})}{2^{2020}}=\dfrac{-1+2^{2020}}{2^{2019}}$
