Question

cho A =\(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\). Chứng tỏ A < 1 

Answer 1

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

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

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

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

Vì \(1-\frac{1}{2^{100}}\)< 1

Nên \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\) < 1

Answer 2

Em làm lại bài đó rồi, anh đừng nói thế

Answer 3

2A = 2(1/2+1/2^2+1/2^3+1/2^4+...+1/2^100)

     = 1+1/2+1/2^2+1/2^3+...+1/2^99

=> 2A-A = (1+1/2+1/2^2+1/2^3+...+1/2^99)-(1/2+1/2^2+1/2^3+...+1/2^100)

=> 2A-A = 1+1/2+1/2^2+1/2^3+...+1/2^99-1/2+1/2^2+1/2^3+...+1/2^100

=>      A = 1-1/2^100<1

=> ĐIỀU PHẢI CHỨNG MINH