Question

Chứng ming rằng : \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}<1\)

Answer 1

Ta có :

\(\frac{1}{2^2}=\frac{1}{2.2}<\frac{1}{1.2}=\frac{1}{1}-\frac{1}{2}\)

\(\frac{1}{3^2}=\frac{1}{3.3}<\frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\)         

  \(\frac{1}{4^2}=\frac{1}{4.4}<\frac{1}{3.4}=\frac{1}{3}-\frac{1}{4}\)

.......................

\(\frac{1}{100^2}=\frac{1}{100.100}<\frac{1}{99.100}=\frac{1}{99}-\frac{1}{100}\)

Cộng vế với vế , ta được :

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

Vì 99 < 100 nên \(\frac{99}{100}<1\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}<1\left(đpcm\right)\)

Answer 2

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+...+1/100^2< 1-1/2+1/2-1/3+....+1/99-1/100=1-1/100 

=>1/2^2 +1/3^2+...+1/100^2<1
=> ĐPCM

Answer 3

Ta có:\(\frac{1}{2^2}<\frac{1}{1.2};\frac{1}{3^2}<\frac{1}{2.3};............;\frac{1}{100^2}<\frac{1}{99.100}\)

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

Mà \(\frac{1}{1.2}+\frac{1}{2.3}+.............+\frac{1}{99.100}\)

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

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

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

Answer 4

\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
\(<\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(<1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(<1-\frac{1}{100}\)
\(<\frac{99}{100}\)
\(=>\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}<\frac{99}{100}<1\left(đpcm\right)\)

Answer 5

=1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100

=1-1/100

=99/100