Question

Cho A=\(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}\).So sánh A với 1

Answer 1

\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}\)

\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}\)

\(\Rightarrow2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}\right)\)

\(\Rightarrow A=1-\frac{1}{2^{2016}}<1\)

Answer 2

mình mới học lớp 6 à

Answer 3

2A=1+1/2+1/2²+...+1/2²⁰¹⁵

Suy ra A=1-1/2²⁰¹⁶

Suy ra A<1