b) A=\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)

   3A=\(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)

3A-A=\(1-\frac{1}{3^{99}}\)

   2A=\(1-\frac{1}{3^{99}}\)

vì 2A<1

=> A<\(\frac{1}{2}\)

anh làm cho e câu a nữa được không ạ

VIẾT SAI ĐỀ BÀI NHÉ

50<A<100

A=\(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{49\cdot50}\)

\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{49}-\frac{1}{50}\)

A=\(1-\frac{1}{50}\)

\(A=\frac{49}{50}\)

BC chịu thua

Ta có \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\)

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

               \(\frac{1}{2}-\frac{1}{100}=\frac{49}{100}< \frac{3}{4}\left(đpcm\right)\)

Ta có : \(A=\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+......+\frac{1}{2^{100}}\)

\(\Rightarrow4A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^4}+.....+\frac{1}{2^{98}}\)

\(\Rightarrow4A-A=\frac{1}{2}-\frac{1}{2^{100}}\)

\(\Rightarrow3A=\frac{2^{99}-1}{2^{100}}\)

\(\Rightarrow A=\frac{2^{99}-1}{\frac{2^{200}}{3}}\)

Vì : \(\frac{2^{99}-1}{2^{200}}< 1\)

Nên : \(A< \frac{1}{3}\)

Ta thấy : \(\frac{1}{2^2}< \frac{1}{3}\)

             \(\frac{1}{2^4}< \frac{1}{3}\)

                 ...

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

\(\Rightarrow A=\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+...+\frac{1}{2^{100}}< \frac{1}{3}\)

Vậy \(A< \frac{1}{3}\)

Chúc bạn học tốt :>

A.\(4\)=\(1+\frac{1}{2^2}+\frac{1}{2^4}+...+\frac{1}{2^{98}}\)

=> 4A-A=1-\(\frac{1}{2^{100}}\)

=> A=\(\frac{1}{3}\left(1-\frac{1}{2^{100}}\right)=\frac{1}{3}-\frac{1}{3}.\frac{1}{2^{100}}< \frac{1}{3}\)