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Question

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A= 1+1/2+1/2^+1/2^+....+1/2^2017

Nguyễn Thị Huyền Trang · Accepted Answer

$A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2017}}$

$\Rightarrow2A=2+1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2016}}$

$\Rightarrow2A-A=\left(2+1+\dfrac{1}{2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2016}}\right)$

$-\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2017}}\right)$

$\Rightarrow A=2-\dfrac{1}{2^{2017}}=\dfrac{2^{2018}-1}{2^{2017}}$Câu trả lời: $A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2017}}$

$\Rightarrow2A=2+1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2016}}$

$\Rightarrow2A-A=\left(2+1+\dfrac{1}{2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2016}}\right)$

$-\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2017}}\right)$

$\Rightarrow A=2-\dfrac{1}{2^{2017}}=\dfrac{2^{2018}-1}{2^{2017}}$

Mashiro Shiina · Answer

$A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2017}}$

$2A=\left(2+1+\dfrac{1}{2}+...+\dfrac{1}{2^{2016}}\right)$

$2A-A=\left(2+1+\dfrac{1}{2}+...+\dfrac{1}{2^{2016}}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2017}}\right)$

$A=2-2^{2017}$Câu trả lời: $A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2017}}$

$2A=\left(2+1+\dfrac{1}{2}+...+\dfrac{1}{2^{2016}}\right)$

$2A-A=\left(2+1+\dfrac{1}{2}+...+\dfrac{1}{2^{2016}}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2017}}\right)$

$A=2-2^{2017}$