bỏ số 2 ở đằng sau đi nhé

Ta có:

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

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

...

\(\frac{1}{100^2}< \frac{1}{99.100}\)

\(\Rightarrow A=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)

\(\Leftrightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

\(\Leftrightarrow A< 1-\frac{1}{100}< 1\left(đpcm\right)\)

giúp mk vs các bạn ưi ! mk đang cần gấp ai nhanh mik tích cho !nhanh nha help me!thank nhìu

Đặt A=\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{2010^2}\)

Ta có: \(\frac{1}{2^2}< \frac{1}{1\cdot2};\frac{1}{3^2}< \frac{1}{2\cdot3};\frac{1}{4^2}< \frac{1}{3\cdot4};....;\frac{1}{2010^2}< \frac{1}{2009\cdot2010}\)

=> A<\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2009\cdot2010}\)

\(\Leftrightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2009}-\frac{2}{2010}\)

\(\Leftrightarrow A< 1-\frac{1}{2010}\)

<=> A<1 (đpcm)

Ta có \(\frac{1}{2^2}< \frac{1}{1.2}\)

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

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

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

...

\(\frac{1}{2010^2}< \frac{1}{2009.2010}\)

Cộng vế các BĐT trên ta được

\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2010^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2009.2010}\)

\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2010^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2009}-\frac{1}{2010}\)

\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2010^2}< 1-\frac{1}{2010}< 1\)

Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2010^2}< 1\)

 ta có 
1/2^2 < 1/(1.2)= 1-1/2 
1/3^2 <1/(2.3)=1/2-1/3 
1/4^2 <1/(3.4)=1/3-1/4 
...... 
1/100^2 < 1/99-1/100 
cộng vế với vế ta được 1/2^2 +1/3^2+...< 1-1/2+1/2-1/3+....+1/99-1/100=1-1/100 
=> 100/100-1/100

=>99/100

tk nha bn

99/100<1 bn nha

Ta có: 

\(\frac{1}{2^2}< \frac{1}{2.1}=\frac{1}{2}-\frac{1}{3}\)

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

\(\frac{1}{100^2}< \frac{1}{99.100}=\frac{1}{99}-\frac{1}{100}\)

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

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

Đáp số : ....

\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{2009^2}+\frac{1}{2010^2}>1\)

=\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.....+\frac{1}{2009^2}+\frac{1}{2010^2}>\frac{ }{ }\)\(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{2008.2009}+\frac{1}{2009.2010}\)

=\(\frac{1}{1}-\frac{1}{2010}=\frac{2010}{2010}-\frac{1}{2010}\)=\(\frac{2010}{2010}>\frac{1}{2010}=1>\frac{1}{2010}\)

Vậy \(1>\frac{1}{2010}\)

Bạn ơi sai đề nhé

1-1/2+1/3-1/4+...+1/199-1/200

=(1+1/3+...+1/199)-(1/2+1/4+...+1/200)

=(1+1/2+1/3+...+1/199+1/200)-2(1/2+1/4+...+1/200)

=(1+1/2+1/3+...+1/199+1/200)-(1+1/2+...+1/100)

=1/101+1/102+...+1/200 (đpcm)

sửa đề : \(\frac{9}{10!}+\frac{10}{11!}+\frac{11}{12!}+...+\frac{99}{100!}\)

\(=\frac{10-1}{10!}+\frac{11-1}{11!}+\frac{12-1}{12!}+...+\frac{100-1}{100!}\)

\(=\frac{1}{9!}-\frac{1}{10!}+\frac{1}{10!}-\frac{1}{11!}+\frac{1}{11!}-\frac{1}{12!}+...+\frac{1}{99!}-\frac{1}{100!}\)

\(=\frac{1}{9!}-\frac{1}{100!}< \frac{1}{9!}\left(đpcm\right)\)

Ta có \(\frac{1}{k^2}=\frac{4}{4k^2}< \frac{4}{4k^2-1}=2\left(\frac{1}{2k-1}-\frac{1}{2k+1}\right)\left(k\in N\cdot\right)\)

Khi đó \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}< 2\left(\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2n-1}-\frac{1}{2n+1}\right)\\ =2\left(\frac{1}{3}-\frac{1}{2n+1}\right)< \frac{2}{3}\)