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 \(\Rightarrow B< 6\)

a) Ta có : \(\frac{1}{2^2}< \frac{1}{1\cdot2}\)

\(\frac{1}{4^2}< \frac{1}{3\cdot4}\)

. . .

\(\frac{1}{100^2}< \frac{1}{99\cdot100}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{2^2}\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{49\cdot50}\right)\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{4}\left(1+1-\frac{1}{50}\right)\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{4}\cdot\frac{99}{50}=\frac{99}{200}< \frac{100}{200}=\frac{1}{2}\left(đpcm\right)\)

b) Ta có :

\(B=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{2499}{2500}>48\)

\(\Rightarrow1-\frac{1}{4}+1-\frac{1}{9}+...+1-\frac{1}{2500}>48\)

\(\Rightarrow49-\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)< 49\)

Lại có : \(\frac{1}{2^2}< \frac{1}{1\cdot2}\)

\(\frac{1}{3^2}< \frac{1}{2\cdot3}\)

. . .

\(\frac{1}{50^2}< \frac{1}{49\cdot50}\)

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

\(\Rightarrow\frac{1}{2^2}+...+\frac{1}{50^2}< \frac{49}{50}< 1\)

\(\Rightarrow-\left(\frac{1}{2^2}+...=\frac{1}{50^2}\right)>1\)

\(\Rightarrow49-\left(\frac{1}{2^2}+...+\frac{1}{50^2}\right)>49-1=48\)

hay \(\frac{3}{4}+\frac{8}{9}+...+\frac{2499}{2500}>48\left(đpcm\right)\)

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

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

\(A< \frac{1}{2^2}.\left(1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\right)\)

\(A< \frac{1}{4}.\left(1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\right)\)

\(A< \frac{1}{4}.\left(2-\frac{1}{50}\right)< \frac{1}{4}.2=2\)

=> \(A< 2\left(đpcm\right)\)

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

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

\(A< \frac{1}{2^2}.\left(1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\right)\)

\(A< \frac{1}{4}.\left(1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\right)\)

\(A< \frac{1}{4}.\left(2-\frac{1}{50}\right)< \frac{1}{4}.2=2\)

\(A< 2\left(đpcm\right)\)

\(\Rightarrow\)A<2(dpcm)

A=1/4^2+1/6^2+1/8^2+....+1/100^2=1/16+1/36+1/64+...+1/10000<1/15+1/35+1/63+...+1/9999

Ta có :1/15+1/35+1/63+...+1/9999=1/3.5+1/5.7+1/7.9+...+1/99.101=1/2.(2/3.5+2/5.7+2/7.9+...+2/99.101)

=1/2.(1/3-1/5+1/5-1/7+1/7-1/9+...+1/99-1/101)=1/2.(1/3-1/101)=1/2.100/101=100/202=50/101

bạn Hoàng ơi ban giải bị sai rồi kìa

\(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}=\frac{1}{2^2}.\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)\)

\(>\frac{1}{2^2}.\left(1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4.5}+...+\frac{1}{50.51}\right)=\frac{1}{4}.\left(1+\frac{1}{4}+\frac{1}{9}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{50}-\frac{1}{51}\right)\)

\(=\frac{1}{4}.\left(1+\frac{1}{4}+\frac{1}{4}+\frac{1}{9}-\frac{1}{51}\right)>\frac{1}{4}.\left(1+\frac{1}{4}+\frac{1}{4}+\frac{1}{9}-\frac{1}{9}\right)=\frac{1}{4}.\left(1+\frac{1}{4}+\frac{1}{4}\right)=\frac{1}{4}.\frac{3}{2}=\frac{3}{8}\)

\(\Rightarrow A>\frac{3}{8}\left(đpcm\right)\)

cảm ơn bạn nhé

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

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