nhầm tớ lộn sang bài khác sorry

trình bày cách giải giùm với nhé

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

=>\(\frac{1}{3}.A=\frac{1}{3^2}-\frac{2}{3^3}+\frac{3}{3^4}-\frac{4}{3^5}+...+\frac{99}{3^{100}}-\frac{100}{3^{101}}\)

=>\(A+\frac{1}{3}.A=\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}+\frac{1}{3^2}-\frac{2}{3^3}+\frac{3}{3^4}-\frac{4}{3^5}+...+\frac{99}{3^{100}}-\frac{100}{3^{101}}\)

=>\(\frac{4}{3}.A=\frac{1}{3}-\left(\frac{2}{3^2}-\frac{1}{3^2}\right)+\left(\frac{3}{3^3}-\frac{2}{3^3}\right)-\left(\frac{4}{3^4}-\frac{3}{3^4}\right)+...+\left(\frac{99}{3^{99}}-\frac{98}{3^{99}}\right)-\left(\frac{100}{3^{100}}-\frac{99}{3^{100}}\right)-\frac{100}{3^{101}}\)

=>\(\frac{4}{3}.A=\frac{1}{3}-\frac{1}{3^2}+\frac{1}{3^3}-\frac{1}{3^4}+...+\frac{1}{3^{99}}-\frac{1}{3^{100}}-\frac{100}{3^{101}}\)

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

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

=>\(B+\frac{1}{3}.B=\frac{1}{3}-\frac{1}{3^2}+\frac{1}{3^3}-\frac{1}{3^4}+...+\frac{1}{3^{99}}-\frac{1}{3^{100}}+\frac{1}{3^2}-\frac{1}{3^3}+\frac{1}{3^4}-\frac{1}{3^5}+...+\frac{1}{3^{100}}-\frac{1}{3^{101}}\)

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

=>\(B=\frac{1}{3}:\frac{4}{3}-\frac{1}{3^{101}}:\frac{4}{3}\)

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

=>\(B=\frac{1}{4}-\frac{1}{3^{100}.4}\)

Lại có: \(\frac{4}{3}.A=\frac{1}{3}-\frac{1}{3^2}+\frac{1}{3^3}-\frac{1}{3^4}+...+\frac{1}{3^{99}}-\frac{1}{3^{100}}-\frac{100}{3^{101}}\)

=>\(\frac{4}{3}.A=B-\frac{100}{3^{101}}\)

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

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

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

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

=>\(\frac{4}{3}.A=\frac{1}{2}-\frac{1}{3^{100}}.\frac{403}{12}\)

Ta thấy: \(\frac{1}{3^{100}}.\frac{403}{12}<\frac{1}{3}.\frac{9}{12}=\frac{1}{3}.\frac{3}{4}=\frac{1}{4}\)

=>\(\frac{1}{3^{100}}.\frac{403}{12}<\frac{1}{4}\)

=>\(\frac{4}{3}.A=\frac{1}{2}-\frac{1}{3^{100}}.\frac{403}{12}<\frac{1}{2}-\frac{1}{4}=\frac{1}{4}\)

=>\(\frac{4}{3}.A<\frac{1}{4}=>A<\frac{1}{4}:\frac{4}{3}=>A<\frac{3}{16}\)

=>\(A<\frac{3}{16}\)

Vậy \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}<\frac{3}{16}\)

=))

Dài quá bạn ơi!!!

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