Question

So sánh : A=\(\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{50}}\)với 1

Answer 1

Ta có:

\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+....+\dfrac{1}{2^{50}}\)

\(\Rightarrow2A=2\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+....+\dfrac{1}{2^{50}}\right)\)

\(\Rightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+....+\dfrac{1}{2^{49}}\)

\(\Rightarrow2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+....+\dfrac{1}{2^{49}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+....+\dfrac{1}{2^{50}}\right)\)

\(\Rightarrow A=1-\dfrac{1}{50}< 1\)

Vậy A < 1