Question

Cho S = \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2018}}\)   

Chứng tỏ rằng S < 1

Answer 1

\(S=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2018}}\)

 \(2S=2.\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2018}}\right)\)

\(2S=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2017}}\)

\(2S-S=\left(1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{2017}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^{2018}}\right)\)

\(S=1-\frac{1}{2^{2018}}< 1\)

Answer 2

\(S=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{...1}{2^{2018}}\)

\(\Rightarrow2S=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2017}}\)
\(2S-S=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2017}}-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{...1}{2^{2018}}\right)\)
\(S=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2017}}-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{2018}}\)
\(S=1-\frac{1}{2^{2018}}\)
\(Mà 1-\frac{1}{2^{2018}}< 1\)
\(\Rightarrow S< 1\)

Answer 3

Nguyễn Phúc Hậu Đã trổ tài r đó, 

Answer 4

2S=1+1/2+1/2^2+....+1/2^2017

TÍNH ĐỂ RA:

2S-S=1-1/2^2018

SUY RA: 

S=1-1/2^2018

MÀ 1-1/2^2018<1

SUY RA  :S<1(ĐPCM)