Question

So sánh A và B, biết:

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

và\(B=1\)

Help me!

Answer 1

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

\(2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}\)

\(2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\right)\)

\(A=1-\dfrac{1}{2^{100}}< 1\)

Vậy A < B.

Answer 2

Giải:

Có: \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\)

\(\Leftrightarrow\dfrac{1}{2}A=\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{101}}\)

Lấy vế trừ theo vế, ta được:

\(A-\dfrac{1}{2}A=\dfrac{1}{2}A=\dfrac{1}{2}-\dfrac{1}{2^{101}}\)

\(\Leftrightarrow\dfrac{1}{2}A=\dfrac{1}{2}-\dfrac{1}{2^{101}}\)

\(\Leftrightarrow A=\dfrac{\dfrac{1}{2}-\dfrac{1}{2^{101}}}{\dfrac{1}{2}}\)

\(\Leftrightarrow A=\dfrac{\dfrac{1}{2}\left(1-\dfrac{1}{2^{100}}\right)}{\dfrac{1}{2}}\)

\(\Leftrightarrow A=1-\dfrac{1}{2^{100}}\)

Lại có \(B=1\)

Vì \(1-\dfrac{1}{2^{100}}< 1\)

Nên \(A< B\)

Vậy \(A< B\).

Chúc bạn học tốt!

Answer 3

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

\(2A=2\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\right)\)

\(2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}\)

\(2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\right)\)\(A=1-\dfrac{1}{2^{100}}\)

\(A< 1\Rightarrow A< B\)