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Đặt \(A=\frac{1}{5}+\frac{1}{5^3}+...+\frac{1}{5^{101}}\)
\(\Rightarrow25A=5+\frac{1}{5}+\frac{1}{5^3}+...+\frac{1}{5^{99}}\)
\(\Rightarrow25A-A=\left(5+\frac{1}{5}+\frac{1}{5^3}+...+\frac{1}{5^{99}}\right)-\left(\frac{1}{5}+\frac{1}{5^3}+\frac{1}{5^5}+...+\frac{1}{5^{101}}\right)\)
hay \(24A=5-\frac{1}{5^{101}}\)
\(\Rightarrow A=\frac{5-\frac{1}{5^{101}}}{24}\)
\(\Rightarrow A:\left(1-\frac{1}{5^{102}}\right)=\frac{5-\frac{1}{5^{101}}}{24}.\frac{1}{1-\frac{1}{5^{102}}}\)
\(=\frac{5\left(1-\frac{1}{5^{102}}\right)}{24}.\frac{1}{1-\frac{1}{5^{102}}}=\frac{5}{24}\)
\(A=\frac{1}{5}+\frac{1}{5^3}+\frac{1}{5^5}+...+\frac{1}{5^{101}}\)
\(\frac{1}{5^2}A=\frac{1}{5^3}+\frac{1}{5^5}+\frac{1}{5^7}+...+\frac{1}{5^{103}}\)
\(\left(1-\frac{1}{5^2}\right)A=\left(\frac{1}{5}+\frac{1}{5^3}+\frac{1}{5^5}+...+\frac{1}{5^{101}}\right)-\left(\frac{1}{5^3}+\frac{1}{5^5}+\frac{1}{5^7}+...+\frac{1}{5^{103}}\right)\)
\(\frac{24}{25}A=\frac{1}{5}-\frac{1}{5^{103}}\)
\(A=\left(1-\frac{1}{5^{102}}\right).\frac{5}{24}\)
Suy ra \(\left(\frac{1}{5}+\frac{1}{5^3}+\frac{1}{5^5}+...+\frac{1}{5^{101}}\right)\div\left(1-\frac{1}{5^{102}}\right)=\frac{5}{24}\).
Đặt S = \(\frac{1}{2}+\frac{1}{2^5}+\frac{1}{2^9}+...+\frac{1}{2^{101}}\)
=> 24S = 16S = \(2^3+\frac{1}{2}+\frac{1}{2^5}+...+\frac{1}{2^{97}}\)
=> 16S - S = \(2^3+\frac{1}{2}+\frac{1}{2^5}+...+\frac{1}{2^{97}}-\left(\frac{1}{2}+\frac{1}{2^5}+\frac{1}{2^9}+...+\frac{1}{2^{101}}\right)\)
=> 15S = \(2^3-\frac{1}{2^{101}}\)
=> S = \(\frac{2^3-\frac{1}{2^{101}}}{15}\)
Khi đó A = \(\frac{2^3-\frac{1}{2^{101}}}{15}:\left(2^3-\frac{1}{2^{101}}\right)=\frac{1}{15}\)
Đặt S = \(\frac{1}{7^2}+\frac{1}{7^4}+\frac{1}{7^6}+...+\frac{1}{7^{100}}\)
=> 72S = 49S = \(1+\frac{1}{7^2}+\frac{1}{7^4}+...+\frac{1}{7^{98}}\)
=> 49S - S = \(\left(1+\frac{1}{7^2}+\frac{1}{7^4}+...+\frac{1}{7^{98}}\right)-\left(\frac{1}{7^2}+\frac{1}{7^4}+\frac{1}{7^6}+...+\frac{1}{7^{100}}\right)\)
=> 48S = \(1-\frac{1}{7^{100}}\)
=> \(S=\frac{1-\frac{1}{7^{100}}}{48}\)
Khi đó A = \(\left(\frac{1-\frac{1}{7^{100}}}{48}\right):\left(1-\frac{1}{7^{100}}\right)=\frac{1}{48}\)
Đặt S = \(\frac{1}{6}+\frac{1}{6^2}+\frac{1}{6^3}+...+\frac{1}{6^{100}}\)
=> 6S = \(1+\frac{1}{6}+\frac{1}{6^2}+...+\frac{1}{6^{99}}\)
=> 6S - S = \(\left(1+\frac{1}{6}+\frac{1}{6^2}+\frac{1}{6^3}+...+\frac{1}{6^{99}}\right)-\left(\frac{1}{6}+\frac{1}{6^2}+\frac{1}{6^3}+...+\frac{1}{6^{100}}\right)\)
=> \(5S=1-\frac{1}{6^{100}}\)
=> \(S=\frac{1-\frac{1}{6^{100}}}{5}\)
Khi đó A = \(\left(1-\frac{1}{6^{100}}\right):\left(\frac{1-\frac{1}{6^{100}}}{5}\right)=5\)
\(-4\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{6}\right)\le x\le-\frac{2}{3}.\left(\frac{1}{3}-\frac{1}{2}-\frac{3}{4}\right)\)
\(\Rightarrow-\frac{13}{3}.\left(\frac{3}{6}-\frac{1}{6}\right)\le x\le-\frac{2}{3}.\left(\frac{4}{12}-\frac{6}{12}-\frac{9}{12}\right)\)
\(\Rightarrow-\frac{13}{3}.\frac{2}{6}\le x\le-\frac{2}{3}.\frac{-11}{12}\)
\(\Rightarrow\frac{-13}{9}\le x\le\frac{11}{18}\)
\(\Rightarrow\frac{-26}{18}\le x\le\frac{11}{18}\)
=> -1,44444444444........... ≤ x ≤ 0,6111111111...........
Mà x ∈ Z
=> x ∈ { -1 ; 0 }