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

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

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

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

\(\Rightarrow S< 1\left(đpcm\right)\)

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 : \(S< 1.\)

Bài làm

\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{20^2}\)

Ta có: \(\frac{1}{2.2}< \frac{1}{1.2};\frac{1}{3.3}< \frac{1}{2.3};\frac{1}{4.4}< \frac{1}{3.4};\frac{1}{20.20}< \frac{1}{19.20}\)

=> \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{20^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{19.20}\)                               (1)

Biến đổi vế phải, ta có:

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{19.20}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{19}-\frac{1}{20}\)

\(=\frac{1}{1}-\frac{1}{20}\)

\(=\frac{20}{20}-\frac{1}{20}\)

\(=\frac{19}{20}\)

Mà 19 < 20

=> \(\frac{19}{20}< 1\)                                                           (2)

Từ (1) và (2) => \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{20^2}< 1\)

Vậy A < 1 ( đpcm ) 

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 = 1/ 2 + 1/ 2^2 + 1/ 2^3 + ... + 1/ 2^20

Nên 2S = 1 + 1/2 + 1 / 2^2 + 1/ 2^3 + .... + 1/ 2^19

Do đó 2S - S = 1 - 1/ 2^20 < 1 

Vậy S < 1

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

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

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

S<1

sửa đề câu 1 :

\(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{99}{100!}\)

\(=\frac{2-1}{2!}+\frac{3-1}{3!}+\frac{4-1}{4!}+...+\frac{100-1}{100!}\)

\(=\frac{1}{1!}-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+...+\frac{1}{99!}-\frac{1}{100!}\)

\(=1-\frac{1}{100!}< 1\)

sửa đề câu 2

\(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}\)

\(=\frac{1.2}{2!}-\frac{1}{2!}+\frac{2.3}{3!}-\frac{1}{3!}+\frac{3.4}{4!}-\frac{1}{4!}+...+\frac{99.100}{100!}-\frac{1}{100!}\)

\(=\left(\frac{1.2}{2!}+\frac{2.3}{3!}+\frac{3.4}{4!}+...+\frac{99.100}{100!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{100!}\right)\)

\(=\left(1+1+\frac{1}{2!}+...+\frac{1}{98!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{100!}\right)\)

\(=2-\frac{1}{99!}-\frac{1}{100!}< 2\)

khi cộng cac số có tử bé hơn mẫu thì tổng sẽ <1 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