Ta có :

\(C=\frac{1}{4}+\frac{1}{4^2}+.....+\frac{1}{4^{1000}}\)

\(\Rightarrow4C=1+\frac{1}{4}+.....+\frac{1}{4^{1999}}\)

\(\Rightarrow4C-C=\left(1+\frac{1}{4}+.....+\frac{1}{4^{1999}}\right)-\left(\frac{1}{4}+\frac{1}{4^2}+.....+\frac{1}{4^{1000}}\right)\)

\(\Rightarrow3C=1-\frac{1}{4^{1000}}\)

\(\Rightarrow C=\frac{1}{3}-\frac{1}{3.4^{1000}}< \frac{1}{3}\)

=> C < 1 / 3

Ta có:

\(C=\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{1000}}\)

\(\Rightarrow4C=1+\frac{1}{4}+...+\frac{1}{4^{999}}\)

\(\Rightarrow4C-C=\left(1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{999}}\right)-\left(\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{999}}+\frac{1}{4^{1000}}\right)\)

\(\Rightarrow3C=1-\frac{1}{4^{1000}}\)

\(\Rightarrow C=\left(1-\frac{1}{4^{1000}}\right).\frac{1}{3}\)

\(\Rightarrow C=\frac{1}{3}-\frac{1}{4^{1000}.3}\)

Mà \(\frac{1}{3}>\frac{1}{3}-\frac{1}{4^{1000}.3}\)

\(\Rightarrow C< \frac{1}{3}\)

Vậy \(C< \frac{1}{3}\)

\(M=\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{1000}}\)

\(4M=\frac{4}{4}+\frac{4}{4^2}+...+\frac{4}{4^{1000}}\)

\(4M=1+\frac{1}{4}+\frac{1}{4^2}+..+\frac{1}{4^{999}}\)

\(4M-M=\left(1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{999}}\right)-\left(\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{1000}}\right)\)

\(3M=1-\frac{1}{4^{1000}}\)

\(M=\left(1-\frac{1}{4^{1000}}\right):3\)

\(M=\frac{4^{1000}-1}{4^{1000}}:3\)

\(M=\frac{4^{1000}-1}{3.4^{1000}}\)

\(\frac{1}{3}=\frac{4^{1000}}{3.4^{1000}}\)

vì \(\frac{4^{1000}-1}{4^{1000}}< \frac{4^{1000}}{3.4^{1000}}\)

nên \(M< \frac{1}{3}\)

<or>or=<or>=

Sửa đề : \(A=\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+......+\frac{1}{2^{199}}\)

\(\Rightarrow2A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+......+\frac{1}{2^{198}}\)

\(\Rightarrow2A-A=A=\frac{1}{2}-\frac{1}{2^{199}}< \frac{1}{2}+\frac{1}{4}=\frac{3}{4}\)

Vậy \(A< \frac{3}{4}\)

1/100 hả e hay là 1/10

Dạ 1/100

theo đây mà làm giờ a bận chuẩn bị KT 1T r

link:  https://olm.vn/hoi-dap/question/148148.html

\(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}.....\frac{899}{30^2}\)

\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{29.31}{30.30}=\frac{1.2.3.....29}{2.3.4.....30}.\frac{3.4.5.....31}{2.3.4.....30}\)

\(=\frac{1}{2}.\frac{31}{30}=\frac{31}{60}\)

\(2A=1+\frac{1}{2}+...+\frac{1}{2^{49}}\)

\(2A-A=1-\frac{1}{2^{50}}\)

\(A=1-\frac{1}{2^{50}}\)=> A bé hơn 1

tương tự nha

\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\)

\(2A=2.\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\right)\)

\(2A=1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{48}}+\frac{1}{2^{49}}\)

\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{48}}+\frac{1}{2^{49}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\right)\)

\(A=1-\frac{1}{2^{50}}< 1\)

B = \(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}...+\frac{1}{1+2+3+...+2019}\)

    = \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{2019\times1010}\)

    = \(2\times\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{2019\times2020}\right)\)

   = \(2\times\left(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+...+\frac{1}{2019\times2020}\right)\)

  = \(2\times\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2019}-\frac{1}{2020}\right)\)

  = \(2\times\left(\frac{1}{2}-\frac{1}{2020}\right)\)

\(=2\times\frac{1009}{2020}\)

\(=\frac{1009}{1010}< \frac{1010}{1010}=1\)

\(\Rightarrow B< 1\)