A=1+1/2 +1/2 ²+1/2 ³+….+1/2 ²019 rút gọn biểu thức 29/09/2021 Bởi Arianna A=1+1/2 +1/2 ²+1/2 ³+….+1/2 ²019 rút gọn biểu thức
Đáp án: Giải thích các bước giải: $A=1+\dfrac{1}{2}+\dfrac{1}{2^{2}}+…+\dfrac{1}{2^{2019}}$ $ $ $⇒\dfrac{1}{2}.A=\dfrac{1}{2}+\dfrac{1}{2^{2}}+\dfrac{1}{2^{3}}+…+\dfrac{1}{2^{2020}}$ $ $ $⇒\dfrac{1}{2}.A-A=(\dfrac{1}{2}+\dfrac{1}{2^{2}}+\dfrac{1}{2^{3}}+…+\dfrac{1}{2^{2020}})-(1+\dfrac{1}{2}+\dfrac{1}{2^{2}}+…+\dfrac{1}{2^{2019}})$ $ $ $⇒\dfrac{-1}{2}A=\dfrac{1}{2^{2020}}-1$ $ $ $⇒\dfrac{-1}{2}.A=\dfrac{1-2^{2020}}{2^{2020}}$ $ $ $⇒A=\dfrac{-2.(1-2^{2020})}{2^{2020}}=\dfrac{-1+2^{2020}}{2^{2019}}$ Bình luận
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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$⇒A=\dfrac{-2.(1-2^{2020})}{2^{2020}}=\dfrac{-1+2^{2020}}{2^{2019}}$