A=1/(2x2)+1/(3x3)+...+1/(100x100) 
Nhận thấy rằng n x n -1=n x n -n+n-1=n x (n-1)+n-1=(n-1) x (n+1) 
=> A < 1/(2x2-1)+1/(3x3-1)+...+1/(100x100-1)=1/(1x3)+1/(3x5)+...+1/(99x101)=1/2-1/202<1/2<3/4

A=1/(2x2)+1/(3x3)+...+1/(100x100) Nhận thấy rằng n x n -1=n x n -n+n-1=n x (n-1)+n-1=(n-1) x (n+1) => A < 1/(2x2-1)+1/(3x3-1)+...+1/(100x100-1)=1/(1x3)+1/(3x5)+...+1/(99x101)=1/2-1/202<1/2<3/4

la nho hon

Ta có: \(K=\frac14+\frac19+\frac{1}{16}+\cdots+\frac{1}{10000}\)

=>\(K=\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{100^2}\)

=>\(K<\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdots+\frac{1}{99\cdot100}\)

=>\(K<1-\frac12+\frac12-\frac13+\cdots+\frac{1}{99}-\frac{1}{100}\)

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

=>K<1

ra ba\o nhyieu

\(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+.....+\frac{1}{10000}=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+.....+\frac{1}{100.100}\)

\(\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+....+\frac{1}{100.100}<\frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{99.100}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-....-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)\(=1-\frac{1}{100}=\frac{99}{100}<1\)

Vậy \(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+.....+\frac{1}{10000}<1\)

  zxzXZda

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