Question

Chứng minh: 1/2^2 + 1/2^3 + 1/2^4 + ... + 1/2^n < 1

Answer 1

Đặt A=\(\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^n}\)

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

2A-A=\(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{n-1}}-\)\(\left(\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^n}\right)\)

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

Vì \(\frac{1}{2}-\frac{1}{2^n}\) < \(\frac{1}{2}\)

Mà \(\frac{1}{2}\) < 1

Nên \(\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^n}\) < 1

=> đpcm

Answer 2

Đặt A=12² +12³ +12⁴ +...+12ⁿ 

2A=12 +12² +12³ +...+12ⁿ⁻¹ 

2A-A=12 +12² +12³ +...+12ⁿ⁻¹ −(12² +12³ +12⁴ +...+12ⁿ )

A=12 −12ⁿ 

Vì 12 −12ⁿ  < 12 

Mà 12  < 1

Nên 12² +12³ +12⁴ +...+12ⁿ  < 1

=> đpcm

Answer 3

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