Question

Cho S=\(\frac{1}{2^0}+\frac{2}{2^1}+\frac{3}{2^2}+...+\frac{1992}{2^{1991}}\). Chứng minh rằng S<4

Answer 1

S = \(\frac{1}{2^0}+\frac{2}{2^1}+\frac{3}{2^2}+...+\frac{1992}{2^{1991}}\)

2.S = \(2+\frac{2}{2^0}+\frac{3}{2^1}+...+\frac{1992}{2^{1990}}\)

=> 2.S - S = \(2+\frac{1}{2^0}+\frac{1}{2^1}+\frac{1}{2^2}+...+\frac{1}{2^{1990}}-\frac{1992}{2^{1991}}\)

=> S = \(2-\frac{1992}{2^{1991}}+\left(\frac{1}{2^0}+\frac{1}{2^1}+\frac{1}{2^2}+...+\frac{1}{2^{1990}}\right)\)

Đặt A = \(\frac{1}{2^0}+\frac{1}{2^1}+\frac{1}{2^2}+...+\frac{1}{2^{1990}}\)

=>2.A = 2 + \(\frac{1}{2^0}+\frac{1}{2^1}+...+\frac{1}{2^{1989}}\)

=> 2.A - A = 2 - \(\frac{1}{2^{1990}}\)=A

Vậy S = \(4-\frac{1}{2^{1990}}-\frac{1992}{2^{1991}}<4\)

 

 

Answer 2

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Answer 3

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Answer 4

\(\text{Ta có : S = }\frac{1}{2^0}+\frac{2}{2^1}+\frac{3}{2^2}+...+\frac{1992}{2^{1991}}\)

\(\Rightarrow\text{2S }=2+\frac{2}{2^0}+\frac{3}{2^1}+...+\frac{1992}{2^{1990}}\)

\(\Rightarrow S =2-\frac{1992}{2^{1991}}+\left(\frac{1}{2^0}+\frac{1}{2^1}+\frac{1}{2^2}+...+\frac{1}{2^{1990}}\right)\)

\(\text{Đặt A = }\frac{1}{2^0}+\frac{1}{2^1}+\frac{1}{2^2}+...+\frac{1}{2^{1990}}\)

\(\Rightarrow2A=2 - \frac{1}{2^0}+\frac{1}{2^1}+...+\frac{1}{2^{1989}}\)

\(\Rightarrow2A - A=2 - \frac{1}{2^{1990}}=A\)

\(\text{Vậy }S=4-\frac{1}{2^{1990}}-\frac{1992}{2^{1991}}< 4\)