Question

chứng minh : \(\dfrac{1}{2^2}\)+\(\dfrac{1}{2^3}\)+\(\dfrac{1}{2^3}\)+.......+\(\dfrac{1}{2^n}\)<1

giúp mink nnhanh nka

Answer 1

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

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

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

Answer 2

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

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

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

A = \(\dfrac{1}{2}-\dfrac{1}{2^n}\)

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

Mà \(\dfrac{1}{2}< 1\)

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

Chúc học tốt!