Đặt \(A=1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2^{2005}}\)
\(A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2005}}\)
\(2A=2\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2005}}\right)\)
\(2A=2+1+\dfrac{1}{2}+...+\dfrac{1}{2^{2004}}\)
\(2A-A=\left(2+1+\dfrac{1}{2}+...+\dfrac{1}{2^{2004}}\right)-\left(1+\dfrac{1}{2}+...+\dfrac{1}{2^{2005}}\right)\)
\(A=2-\dfrac{1}{2^{2005}}\)
Giải:
Ta có: A = \(1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2^{2005}}.\)
= \(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...=\dfrac{1}{2^{2005}}.\)
2A = \(2\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2005}}\right).\)
= \(1+2+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2004}}.\)
2A -A = \(\left(1+2+\dfrac{1}{2}+...+\dfrac{1}{2^{2004}}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2005}}\right).\)
= 2 - \(\dfrac{1}{2^{2005}}.\)
Vậy \(A=2-\dfrac{1}{2^{2005}}.\)
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