ta có 

\(B=1+\left(1-\frac{1}{2}\right)+..+\left(1-\frac{1}{100}\right)\)

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

Vậy A=B

Ta có 

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

\(2A=1+\frac{2}{2}+\frac{3}{2^2}+...+\frac{99}{2^{98}}+\frac{100}{2^{99}}\)

Suy ra \(A=2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\frac{100}{2^{100}}\)

Đặt \(n=\frac{1}{2}\) thì \(A=1+n+n^2+...+n^{99}-\frac{100}{2^{100}}\)

Xét \(B=1+n+n^2+...+n^{99}\Leftrightarrow B.n=n+n^2+n^3+...+n^{100}\)

\(\Leftrightarrow B.n=\left(1+n+n^2+...+n^{99}\right)+\left(n^{100}-1\right)\)

\(\Leftrightarrow B.n=B+n^{100}-1\Leftrightarrow B\left(n-1\right)=n^{100}-1\Leftrightarrow B=\frac{n^{100}-1}{n-1}\)

Suy ra \(A=\frac{\frac{1}{2^{100}}-1}{\frac{1}{2}-1}-\frac{100}{2^{100}}=2\left(1-\frac{1}{2^{100}}\right)-\frac{100}{2^{100}}=-\frac{102}{2^{100}}+2< 2\)

Vậy A < 2

Hong bé ơi.Bé hong follow anh mà đòi xin đáp án của anh à

ko

mình nghĩ là B

1/3²< 1/2.3

1/4²< 1/3.4

...

1/100²< 1/99.100

=> 1/2² + 1/3² + 1/4² + ... + 1/100²< 1/2² + 1/2.3 + 1/3.4 + ... + 1/99.100

A < 1/4 + 1/2 -1/3 + 1/3 - 1/4 +... + 1/99 - 1/100

A < 1/4 + 1/2 -1/100 < 1/4 + 1/2 = 3/4

=> A < 3/4

Ta có: \(\frac{1}{2}A=\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{100}{2^{101}}\)

\(A-\frac{1}{2}A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}-\frac{100}{2^{101}}\)

Ta có: \(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{100}}=1-\frac{1}{2^{100}}< 1\)

\(\Rightarrow\frac{1}{2}A< 1-\frac{100}{2^{101}}\)

\(\Rightarrow A< 2-\frac{200}{2^{101}}< 2\)

Vậy A<2

\(A=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{98}{2^{98}}+\frac{99}{2^{99}}+\frac{100}{2^{100}}\)

\(2A=1+\frac{2}{2}+\frac{3}{2^2}+\frac{4}{2^3}+...+\frac{99}{2^{98}}+\frac{100}{2^{99}}\)

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

Đặt \(B=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}+\frac{1}{2^{99}}\)

\(2B=2+1+\frac{1}{2}+...+\frac{1}{2^{97}}+\frac{1}{2^{98}}\)

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

Do đó: \(A=2-\frac{1}{2^{99}}-\frac{100}{2^{100}}< 2\)