Question

[CHU ANH TUẤN]

cmr:

1/3+2/9+...+100/3¹⁰⁰ <3/4

Answer 1

[I don't need you]

ta có: \(A=\frac{1}{3}+\frac{2}{3^2}+...+\frac{100}{3^{100}}\)

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

\(\Rightarrow A-\frac{1}{3}A=\frac{1}{3}+\left(\frac{2}{3^2}-\frac{1}{3^2}\right)+...+\left(\frac{100}{3^{100}}-\frac{99}{3^{100}}\right)-\frac{100}{3^{101}}\)

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

+) Xét \(B=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}\)

\(\Rightarrow\frac{1}{3}B=\frac{1}{3^2}+\frac{1}{3^3}...+\frac{1}{3^{101}}\)

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

\(\frac{2}{3}B=\frac{1}{3}-\frac{1}{3^{101}}\)

\(\Rightarrow B=\left(\frac{1}{3}-\frac{1}{3^{101}}\right):\frac{2}{3}=\left(\frac{1}{3}-\frac{1}{3^{101}}\right).\frac{3}{2}\)

Thay B vào A, ta có:

\(\frac{2}{3}A=\left(\frac{1}{3}-\frac{1}{3^{101}}\right).\frac{3}{2}-\frac{100}{3^{101}}\)

\(\Rightarrow A=\left(\left(\frac{1}{3}-\frac{1}{3^{101}}\right).\frac{3}{2}-\frac{100}{3^{101}}\right):\frac{2}{3}\)

\(A=\left(\left(\frac{1}{3}-\frac{1}{3^{101}}\right).\frac{3}{2}-\frac{100}{3^{101}}\right).\frac{3}{2}\)

\(A=\frac{9}{4}.\left(\frac{1}{3}-\frac{1}{3^{101}}\right)-\frac{100.3}{3^{101}.2}=\frac{9}{4}.\left(\frac{1}{3}-\frac{1}{3^{101}}\right)-\frac{150}{3^{101}}\)

                                                            \(A=\frac{3}{4}-\frac{9}{4}.\frac{1}{3^{101}}-\frac{150}{3^{101}}< \frac{3}{4}\)

\(\Rightarrow A=\frac{1}{3}+\frac{2}{3^2}+...+\frac{100}{3^{100}}< \frac{3}{4}\left(đpcm\right)\)