Đặt \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2018}}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2017}}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2017}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2018}}\right)\)
\(A=1-\frac{1}{2^{2018}}< 1\)
\(\Rightarrow A< 1\left(đpcm\right)\)
hok tốt .
xin lỗi nha , mk ko thấy S bạn thay A => S là đc
bạn thông cảm ,
S=1/2 + 1/2^2 + 1/2^3+... +1/2^2018
2S=1+1/2^2+1/2^3+...+1/2^2017
2S-S=(1+1/2^2+1/2^3+...+1/2^2017)-(1/2+1/2^2+1/2^3+...+1/2^2018)
S=1 - 1/2^2018 < 1
Vậy S < 1 ( điều phải chứng minh )
\(S=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+..+\frac{1}{2^{2018}}\)
\(\Rightarrow2S=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2017}}\)
\(\Rightarrow2S-S=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2017}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2018}}\right)\)
\(\Rightarrow S=1-\frac{1}{2^{2018}}< 1\)(đpcm)
_Chúc bạn học tốt_
