\(A=\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{8}-\frac{1}{16}\right)+\left(\frac{1}{32}-\frac{1}{64}\right)\)

\(A=\frac{1}{4}+\frac{1}{16}+\frac{1}{64}=\frac{21}{64}<\frac{21}{63}\)

\(\Rightarrow A<\frac{21}{63}=\frac{1}{3}\)

AI tích mk mk sẽ tích lại     

a: Ta có: \(A=1+3^2+3^4+\cdots+3^{100}\)

\(=\left(1+3^2+3^4\right)+\left(3^6+3^8+3^{10}\right)+\cdots+\left(3^{96}+3^{98}+3^{100}\right)\)

\(=\left(1+3^2+3^4\right)+3^6\left(1+3^2+3^4\right)+\cdots+3^{96}\left(1+3^2+3^4\right)=91\left(1+3^6+\cdots+3^{96}\right)\) ⋮91

b: Ta có: \(A=1+3^2+3^4+\cdots+3^{100}\)

=>\(9A=3^2+3^4+3^6+\cdots+3^{102}\)

=>\(9A-A=3^2+3^4+\cdots+3^{102}-1-3^2-\cdots-3^{100}\)

=>\(8A=3^{102}-1\)

=>\(8A+1=3^{102}\)

=>\(8A+1=\left(3^{51}\right)^2\) là số chính phương

Vì \(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}<\frac{1}{12}+\frac{1}{12}+\frac{1}{12}=\frac{3}{12}=\frac{1}{4}\)

\(\frac{1}{61}+\frac{1}{62}+...+\frac{1}{66}<\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}=\frac{6}{60}=\frac{1}{10}\)

=> A < \(\frac{1}{3}+\frac{1}{4}+\frac{1}{10}=\frac{41}{60}<\frac{45}{60}=\frac{3}{4}\)điều phải c/m