Ta đặt: A=\(\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{100}}\)
\(\Rightarrow4A=1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{99}}\)
4A - A = \(1-\frac{1}{4^{100}}\)= 3A = \(\frac{4^{100}-1}{4^{100}}\)\(\Rightarrow A=\frac{4^{100}-1}{4^{100}.3}=\frac{1}{3}-\frac{1}{4^{100}.3}=\frac{1}{3}-\frac{1}{4^{100}}.\frac{1}{3}=\frac{1}{3}\left(1-\frac{1}{4^{100}}\right)\)
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