Question

1.So sánh: A=\(\frac{1}{2^1}+\frac{1}{2^2}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\) và 1:

Answer 1

Tìm 2A 

Rồi lấy 2A - A là ra

Ok

Answer 2

\(A=\frac{1}{2}+\frac{1}{2^2}+.....+\frac{1}{2^{49}}+\frac{1}{2^{50}}\)

\(2A=2.\left(\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{49}}+\frac{1}{2^{50}}\right)\)

\(2A=1+\frac{1}{2}+\frac{1}{2^2}+.....+\frac{1}{2^{49}}\)

\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{48}}+\frac{1}{2^{49}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+.....+\frac{1}{2^{50}}\right)\)

\(A=1-\frac{1}{2^{50}}\)

\(\Rightarrow A< 1\)

Answer 3

bạn ss trung gian 

vd 1/2²<1/1.2