Ta có : \(A=\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+......+\frac{1}{2^{100}}\)

\(\Rightarrow4A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^4}+.....+\frac{1}{2^{98}}\)

\(\Rightarrow4A-A=\frac{1}{2}-\frac{1}{2^{100}}\)

\(\Rightarrow3A=\frac{2^{99}-1}{2^{100}}\)

\(\Rightarrow A=\frac{2^{99}-1}{\frac{2^{200}}{3}}\)

Vì : \(\frac{2^{99}-1}{2^{200}}< 1\)

Nên : \(A< \frac{1}{3}\)

Ta thấy : \(\frac{1}{2^2}< \frac{1}{3}\)

             \(\frac{1}{2^4}< \frac{1}{3}\)

                 ...

              \(\frac{1}{2^{100}}< \frac{1}{3}\)

\(\Rightarrow A=\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+...+\frac{1}{2^{100}}< \frac{1}{3}\)

Vậy \(A< \frac{1}{3}\)

Chúc bạn học tốt :>

A.\(4\)=\(1+\frac{1}{2^2}+\frac{1}{2^4}+...+\frac{1}{2^{98}}\)

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

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

Ta có:\(4A=1+\frac{1}{2^2}+\frac{1}{2^4}+.........+\frac{1}{2^{98}}\)

\(\Rightarrow4A-A=3A=1-\frac{1}{2^{100}}< 1\Rightarrow A< \frac{1}{3}\left(ĐPCM\right)\)

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

Ta có :

\(\frac{1}{2^2}< \frac{1}{1.2}=1-\frac{1}{2}\)

\(\frac{1}{3^2}< \frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\)

\(\frac{1}{4^2}< \frac{1}{3.4}=\frac{1}{3}-\frac{1}{4}\)

\(.......\)

\(\frac{1}{50^2}< \frac{1}{49.50}=\frac{1}{49}-\frac{1}{50}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{49}-\frac{1}{50}=1-\frac{1}{50}< 1\)

\(\Rightarrow1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{50^2}< 2\)

\(\Rightarrow A< \frac{1}{2^2}.2=\frac{1}{2}\) (đpcm)

\(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}\)

\(A< \frac{1}{2^2-1}+\frac{1}{4^2-1}+\frac{1}{6^2-1}+...+\frac{1}{100^2-1}\)

\(A< \frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)

\(2A< \frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)

\(2A< \frac{1}{1}-\frac{1}{101}< 1\Rightarrow2A< 1\Rightarrow A< \frac{1}{2}\left(đpcm\right)\)

\(\frac{1}{2}\)

this sentence extremely easy

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

Đặt B=\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+..+\)\(\frac{1}{99.100}=\)\(1-\frac{1}{100}< 1\)

Mà A=1+B=>A=1+B<1+1=2

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

\(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)

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

\(A=1-\frac{1}{100}\)

vậy \(A=\frac{99}{100}< 2\left(đpcm\right)\)

B)

ta có : \(1=1\)

\(\frac{1}{2}+\frac{1}{3}< \frac{1}{2}+\frac{1}{2}=1\)

\(\frac{1}{4}+\frac{1}{5}+...+\frac{1}{7}< \frac{1}{4}+...+\frac{1}{4}=\frac{4}{4}=1\)

\(\frac{1}{8}+\frac{1}{9}+...+\frac{1}{15}< \frac{1}{8}+...+\frac{1}{8}=\frac{8}{8}=1\)

\(\frac{1}{16}+\frac{1}{17}+...+\frac{1}{63}< 1\)

tất cả công lại \(\Rightarrow B< 6\)