Question

Bài toán : So sánh A với \(\frac{1}{3}\)

\(A=\frac{1}{3^1}-\frac{1}{3^2}+\frac{1}{3^3}-\frac{1}{3^4}+...+\frac{1}{3^{99}}-\frac{1}{3^{100}}\)

Answer 1

#)Giải :

\(A=\frac{1}{3^1}-\frac{1}{3^2}+\frac{1}{3^3}-\frac{1}{3^4}+...+\frac{1}{3^{99}}-\frac{1}{3^{100}}\)

\(A=\frac{2}{9^1}+\frac{2}{9^2}+\frac{2}{9^3}+...+\frac{2}{9^{50}}\)

\(\Rightarrow2A=1+\frac{2}{9}+\frac{2}{9^1}+\frac{2}{9^2}+\frac{2}{9^3}+...+\frac{2}{9^{49}}\)

\(\Rightarrow2A-A=A=\left(1+\frac{2}{9}+\frac{2}{9^1}+\frac{2}{9^2}+\frac{2}{9^3}+...+\frac{2}{9^{49}}\right)-\left(\frac{2}{9^1}+\frac{2}{9^2}+\frac{2}{9^3}+...+\frac{2}{9^{50}}\right)\)

\(\Rightarrow A=1+\frac{2}{9}-\frac{2}{9^{50}}=\frac{11}{9}-\frac{2}{9^{50}}\)

Có lẽ đúng .........................

Answer 2

#)So sánh thì tự làm nhé !

Answer 3

\(A=\frac{1}{3^1}-\frac{1}{3^2}+\frac{1}{3^3}-\frac{1}{3^4}+...+\frac{1}{3^{99}}-\frac{1}{3^{100}}\)

\(\frac{1}{3}A=\frac{1}{3^2}-\frac{1}{3^3}+\frac{1}{3^4}-\frac{1}{3^5}+...+\frac{1}{3^{100}}-\frac{1}{3^{101}}\)

\(\Rightarrow A+\frac{1}{3}A=\frac{1}{3^1}+\left(\frac{-1}{3^{101}}\right)=\frac{1}{3^1}-\frac{1}{3^{101}}\)

\(\Rightarrow A\left(1+\frac{1}{3}\right)=\frac{1}{3^1}-\frac{1}{3^{101}}\)

\(\Rightarrow\frac{4}{3}A=\frac{1}{3^1}-\frac{1}{3^{101}}\)

          \(A=\left(\frac{1}{3^1}-\frac{1}{3^{101}}\right):\frac{4}{3}\)

          \(A=\left(\frac{1}{3^1}-\frac{1}{3^{101}}\right).\frac{3}{4}\)

          \(A=\frac{1}{3^1}.\frac{3}{4}-\frac{1}{3^{101}}.\frac{3}{4}\)

         \(A=\frac{1}{4}-\frac{1}{3^{100}.4}< \frac{1}{4}< \frac{1}{3}\)

\(\Rightarrow A< \frac{1}{3}\)

      Vậy \(A< \frac{1}{3}\)