a)Đặt A= \(\frac{1}{2}\) - \(\frac{1}{4}\) + \(\frac{1}{8}\) - \(\frac{1}{16}\) + \(\frac{1}{32}\) - \(\frac{1}{64}\) => A=\(\frac{1}{2^1}\) - \(\frac{1}{2^2}\) + \(\frac{1}{2^3}\) - \(\frac{1}{2^4}\) + \(\frac{1}{2^5}\) - \(\frac{1}{2^6}\)

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

=> 3A= 1- \(\frac{1}{2^6}\) <1 => A<\(\frac{1}{3}\) => đpcm.

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

=> 3B=1-\(\frac{2}{3}\) + \(\frac{3}{3^2}\) - \(\frac{4}{3^3}\) +...+\(\frac{99}{3^{98}}\) - \(\frac{100}{3^{99}}\)

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

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

=> 3B= 3-1+\(\frac{1}{3}\) - \(\frac{1}{3^2}\) + \(\frac{1}{3^3}\) - \(\frac{1}{3^4}\) +...+ \(\frac{1}{3^{98}}\)

=> 4B= 3-\(\frac{1}{3^{99}}\) <3 => B<\(\frac{3}{4}\) (2)

=> 4A<B<\(\frac{3}{4}\) => A<\(\frac{3}{16}\) => đpcm.

          A = 1/3 + 2/3² + 3/3³ + ... + 100/3¹⁰⁰ 

=>    3A = 1 + 2/3 + 3/3² + ... + 100/3⁹⁹ 

     -    A =       1/3 + 2/3² + ... + 99/3⁹⁹ + 100/ 3¹⁰⁰ 

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=>    2A = 1 + 1/3 + 1/3² + ... + 1/3⁹⁹ + 100/3¹⁰⁰ 

=>    6A = 3 + 1 + 1/3 + ... + 1/3⁹⁸ + 100/3⁹⁹ 

     -  2A = 1 + 1/3 + 1/3² + ... + 1/3⁹⁸ + 1/3⁹⁹ +100/3¹⁰⁰ 

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    4A = 3 - 99/3⁹⁹ - 100/3¹⁰⁰  <  3

=>    4A  <  3

=>    A  <  3/4

Ta có : A = \(\frac{1}{5^2}+\frac{2}{5^3}+\frac{3}{5^4}+...+\frac{99}{5^{100}}\)

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

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

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

=> 20A = \(1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{98}}-\frac{99}{5^{99}}\)

Lấy 20A trừ A ta có : 

20A - A = \(\left(1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{98}}-\frac{99}{5^{99}}\right)-\left(\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{99}}-\frac{99}{5^{100}}\right)\)

16A = \(1-\frac{99}{5^{99}}+\frac{99}{5^{100}}=1+99\left(\frac{1}{5^{100}}-\frac{1}{5^{99}}\right)=1-\frac{99.4}{5^{100}}\)

=> A = \(\frac{1}{16}-\frac{99}{4.5^{100}}< \frac{1}{16}\left(\text{ĐPCM}\right)\)

Ta có :A=\(\frac{1}{5^2}+\frac{2}{5^3}+.....+\frac{99}{5^{100}}\)

          5A=\(\frac{1}{5}+\frac{2}{5^2}+.....+\frac{99}{5^{99}}\)

      5A -A=\(\left(\frac{1}{5}+\frac{2}{5^2}+...+\frac{99}{5^{99}}\right)\)-\(\left(\frac{1}{5^2}+\frac{2}{5^3}+...+\frac{99}{5^{100}}\right)\)

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

Đặt B=\(\frac{1}{5}+\frac{1}{5^2}+.....+\frac{1}{5^{99}}\)

         5B=\(1+\frac{1}{5}+...+\frac{1}{5^{98}}\)

  5B - B =\(\left(1+\frac{1}{5}+...+\frac{1}{5^{98}}\right)\)- \(\left(\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{99}}\right)\)

      4B  =\(1-\frac{1}{5^{99}}\)

 Ta có :4A = B -\(\frac{99}{5^{100}}\)

          16A = 4B -\(\frac{4.99}{5^{100}}\)=\(1-\frac{1}{5^{99}}-\frac{4.99}{5^{100}}\)

              A = \(\frac{1}{16}-\frac{1}{5^{99}.16}-\frac{99}{5^{100}.4}\)< \(\frac{1}{16}\)  

              Suy ra: A <\(\frac{1}{16}\)

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