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B=(3 10/99+4 11/99-58/299).(1/2-4/3-1/6)

=(......................................).0

=0.

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

\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)

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

\(2A=1-\frac{1}{3^{99}}\)

\(A=\frac{1-\frac{1}{3^{99}}}{2}\)

Ta đặt \(C=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

Ta so sánh giữa A và C.

\(\frac{1}{3}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{3^3}< \frac{1}{3.4};....;\frac{1}{3^{99}}< \frac{1}{99.100}\Leftrightarrow A< C\)( 1 )

 \(C=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}=\frac{1}{1}-\frac{1}{100}=\frac{99}{100}\)

Mà \(\frac{99}{100}< \frac{1}{2}\Rightarrow C< B\)( 2 )

Từ ( 1 ) và ( 2 )

 \(\Rightarrow A< C< B\Leftrightarrow A< B\)

A=1/3 +1/3^2+...+1/3^99

-> 3A= 1+1/3+1/3^2+...+1/3^98

-> 3A-A=2A=1-1/3^99 

A=(1-1/3^99)/2. vì 1-1/3^99 < 1  -> A< B.               Vậy A < B

ta có: \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}+\frac{1}{3^{100}}\)

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

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

\(\Rightarrow\frac{2}{3}A< \frac{1}{3}\)

\(\Rightarrow A< \frac{1}{3}:\frac{2}{3}\)

\(\Rightarrow A< \frac{1}{2}\)

a,\(\frac{-3}{1.3}+\frac{-3}{3.5}+....+\frac{-3}{97.99}\)

= -3.\(\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\right)\)

=\(\frac{-3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{97}-\frac{1}{99}\right)\)

=\(\frac{-3}{2}\left(1-\frac{1}{99}\right)\)

=\(\frac{-3}{2}.\frac{98}{99}\)

=\(\frac{49}{-33}\)>\(\frac{49}{-20}\)

Đặt \(K=\frac{1}{3}+\frac{1}{3^2}+.....+\frac{1}{3^{99}}\)

\(3K=1+\frac{1}{3}+\frac{1}{3^2}+.....+\frac{1}{3^{98}}\)

\(3K-K=1+\frac{1}{3}+\frac{1}{3^2}+.....+\frac{1}{3^{98}}-\left(\frac{1}{3}+\frac{1}{3^2}+.....+\frac{1}{3^{99}}\right)\)

\(2K=\)\(1-\frac{1}{3^{99}}\)

\(K=\frac{1-\frac{1}{3^{99}}}{2}\)

Có \(1-\frac{1}{3^{99}}\) < \(\frac{1}{2}\)

\(\Rightarrow K\) < \(\frac{1}{2}\)

Vậy \(\left(\frac{1}{3}+\frac{1}{3^2}+.....+\frac{1}{3^{99}}\right)\) < \(\frac{1}{2}\)

#)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 .........................

#)So sánh thì tự làm nhé !

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

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

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

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

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

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

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

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

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

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

      Vậy \(A< \frac{1}{3}\)