\(S=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{20}}\)

\(\Rightarrow2S=1+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{19}}\)

\(\Rightarrow2S-S=\left(1+\frac{1}{2^2}+...+\frac{1}{2^{19}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{20}}\right)\)

\(S=1-\frac{2}{2^{20}}\)

\(\Rightarrow S< 1\left(đpcm\right)\)

Ta có :\(S=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{20}}\)

\(S=\frac{1\cdot2^{19}}{2\cdot2^{19}}+\frac{1\cdot2^{18}}{2^2\cdot2^{18}}+\frac{1\cdot2^{17}}{2^3\cdot2^{17}}+...+\frac{1\cdot2}{2^{19}\cdot2}+\frac{1}{2^{20}}\)

\(S=\frac{2^{19}}{2^{20}}+\frac{2^{18}}{2^{20}}+\frac{2^{17}}{2^{20}}+...+\frac{2}{2^{20}}+\frac{1}{2^{20}}\)

\(S=\frac{2^{19}+2^{18}+2^{17}+...+2^1+1}{2^{20}}\)

\(S=\frac{2^0+2^1+2^2+...+2^{19}}{2^{20}}\)

Xét: Gọi \(N=2^0+2^1+2^2+...+2^{19}\)

\(2\cdot N=2^1\cdot2^2\cdot2^3\cdot...\cdot2^{20}\)

\(2\cdot N-N=\left(2^1+2^2+2^3+...+2^{20}\right)-\left(2^0+2^1+2^2+...+2^{19}\right)\)

\(N=2^{20}-2^0\)

Thay N vào S, ta có :

\(S=\frac{2^{20}-2^0}{2^{20}}\)

\(S=\frac{2^{20}}{2^{20}}-\frac{1}{2^{20}}\)

\(S=1-\frac{1}{2^{20}}\)

Vì \(1-\frac{1}{2^{20}}< 1\Rightarrow S< 1\left(Đpcm\right).\)

Vậy : \(S< 1.\)

Bạn nhân S với 2

Lấy 2S-S=1-1/(2^20)

S=1/(2^20) nên < 2

Cần làm đầy đủ hơn thì bảo mình

Ta có : 1/2 < 1

             1/2^2 < 1/2

              ..............

                  1/2^19 < 1/2^20

Suy ra 1/2+1/2^2+......+1/2^19<1+1/2+1/2^2+......+1/2^20

Suy ra 1/2+1/2^2+.......+1/2^19+1/2^20<1+1/2+1/2^2+.....+1/2^20+1/2^20

Suy ra S<S+1+1/2^20

Suy ra S<S+1+1/2^20<2

Suy ra S<2

giải chi tiết nha!

ta có 

S = \(\frac{1}{2}\)+ \(\frac{1}{2^2}\)+ \(\frac{1}{2^3}\)+ .....+\(\frac{1}{20^{20}}\)

2S= 1 + \(\frac{1}{2}\)+ \(\frac{1}{2^2}\)+ \(\frac{1}{2^3}\)+ .....+\(\frac{1}{2^{19}}\)

S = 2S-S= 1 - \(\frac{1}{2^{19}}\)<1 

Vậy S < 1 

^_^ chúc bn học tốt

2.a) Vào question 126036

b) Vào question 68660

Nguyen svtkvtm Khôi Bùi Nguyễn Việt Lâm Lê Anh Duy Nguyễn Thành Trương DƯƠNG PHAN KHÁNH DƯƠNG An Võ (leo) Ribi Nkok Ngok Bonking ...

sửa đề : S < 1

\(s< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+..................+\frac{1}{9.10}\)

\(\Leftrightarrow S< 1-\frac{1}{10}\)

vậy S < 1

Đặt A = \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{20}}\)

=> 2A = 1 + \(\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{19}}\)

=> 2A - A = A = 1 - \(\frac{1}{2^{20}}\)<1

\(2A=1+\frac{1}{2}+\frac{1}{2^3}+...+\frac{1}{2^{19}}\)

\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{19}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{20}}\right)\)

\(A=1-\frac{1}{2^{20}}\)

\(vì\) \(1-\frac{1}{2^{20}}< 1\)\(nên\)\(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{20^{20}}< 1\)