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bn ơi???

Ta có: \(S=\frac{1}{5^2}+\frac{2}{5^3}+\cdots+\frac{99}{5^{100}}\)

=>\(5S=\frac15+\frac{2}{5^2}+\cdots+\frac{99}{5^{99}}\)

=>\(5S-S=\frac15+\frac{2}{5^2}+\cdots+\frac{99}{5^{99}}-\frac{1}{5^2}-\frac{2}{5^3}-\cdots-\frac{99}{5^{100}}=\frac15+\frac{1}{5^2}+\frac{1}{5^3}+\cdots+\frac{1}{5^{99}}-\frac{99}{5^{100}}\)

=>\(4S=\frac15+\frac{1}{5^2}+\frac{1}{5^3}+\cdots+\frac{1}{5^{99}}-\frac{99}{5^{100}}\)

Ta có: \(A=\frac{1}{5^2}+\frac{1}{5^3}+\cdots+\frac{1}{5^{99}}\)

=>\(5A=\frac15+\frac{1}{5^2}+\cdots+\frac{1}{5^{98}}\)

=>\(5A-A=\frac15+\frac{1}{5^2}+\cdots+\frac{1}{5^{98}}-\frac{1}{5^2}-\frac{1}{5^3}-\cdots-\frac{1}{5^{99}}\)

=>\(4A=\frac15-\frac{1}{5^{99}}=\frac{5^{98}-1}{5^{99}}\)

=>\(A=\frac{5^{98}-1}{4\cdot5^{99}}\)

Ta có: \(4S=\frac15+\frac{1}{5^2}+\frac{1}{5^3}+\cdots+\frac{1}{5^{99}}-\frac{99}{5^{100}}\)

\(=\frac15+\frac{5^{98}-1}{4\cdot5^{99}}-\frac{99}{5^{100}}=\frac15+\frac{5^{99}-5-396}{4\cdot5^{100}}=\frac15+\frac{1}{4\cdot5}-\frac{401}{4\cdot5^{100}}\)

=>\(4S<\frac15+\frac{1}{20}=\frac{4}{20}+\frac{1}{20}=\frac{5}{20}=\frac14\)

hay S<1/16

Ta có: \(S=\frac{1}{5^2}+\frac{2}{5^3}+\cdots+\frac{99}{5^{100}}\)

=>\(5S=\frac15+\frac{2}{5^2}+\cdots+\frac{99}{5^{99}}\)

=>\(5S-S=\frac15+\frac{2}{5^2}+\cdots+\frac{99}{5^{99}}-\frac{1}{5^2}-\frac{2}{5^3}-\cdots-\frac{99}{5^{100}}=\frac15+\frac{1}{5^2}+\frac{1}{5^3}+\cdots+\frac{1}{5^{99}}-\frac{99}{5^{100}}\)

=>\(4S=\frac15+\frac{1}{5^2}+\frac{1}{5^3}+\cdots+\frac{1}{5^{99}}-\frac{99}{5^{100}}\)

Ta có: \(A=\frac{1}{5^2}+\frac{1}{5^3}+\cdots+\frac{1}{5^{99}}\)

=>\(5A=\frac15+\frac{1}{5^2}+\cdots+\frac{1}{5^{98}}\)

=>\(5A-A=\frac15+\frac{1}{5^2}+\cdots+\frac{1}{5^{98}}-\frac{1}{5^2}-\frac{1}{5^3}-\cdots-\frac{1}{5^{99}}\)

=>\(4A=\frac15-\frac{1}{5^{99}}=\frac{5^{98}-1}{5^{99}}\)

=>\(A=\frac{5^{98}-1}{4\cdot5^{99}}\)

Ta có: \(4S=\frac15+\frac{1}{5^2}+\frac{1}{5^3}+\cdots+\frac{1}{5^{99}}-\frac{99}{5^{100}}\)

\(=\frac15+\frac{5^{98}-1}{4\cdot5^{99}}-\frac{99}{5^{100}}=\frac15+\frac{5^{99}-5-396}{4\cdot5^{100}}=\frac15+\frac{1}{4\cdot5}-\frac{401}{4\cdot5^{100}}\)

=>\(4S<\frac15+\frac{1}{20}=\frac{4}{20}+\frac{1}{20}=\frac{5}{20}=\frac14\)

hay S<1/16

ban h cho minh di

\(S=5\left(\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{100^2}\right)\)Ta có :

 \(S< 5\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)=5\left(1-\frac{1}{100}\right)< 5\)

\(S>5\left(\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{100.101}\right)=5\left(\frac{1}{2}-\frac{1}{101}\right)>2\)

\(\Rightarrow2< S< 5\)

Ta có: \(\frac{1}{2^2}< \frac{1}{1.2}=\frac{1}{1}-\frac{1}{2}\); \(\frac{1}{3^2}< \frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\); ...; \(\frac{1}{100^2}< \frac{1}{99.100}=\frac{1}{99}-\frac{1}{100}\)

=> S < \(5\left(1-\frac{1}{100}\right)=5.\frac{99}{100}< 5.1=5\)=> S<5

Lại có: \(\frac{1}{2^2}>\frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\); \(\frac{1}{3^2}>\frac{1}{3.4}=\frac{1}{3}-\frac{1}{4}\); \(\frac{1}{100^2}>\frac{1}{100.101}=\frac{1}{100}-\frac{1}{101}\)

=> \(S>5\left(\frac{1}{2}-\frac{1}{101}\right)=5.\frac{101-2}{2.101}=\frac{5.99}{2.101}~2,45\)=> S>2

Vậy 2 < S < 5 => Đpcm