\(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}\)

\(\Rightarrow A>\frac{1}{2}.\frac{2}{3}.\frac{4}{5}...\frac{98}{99}\)

\(\Rightarrow A^2>\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}...\frac{98}{99}.\frac{99}{100}\)

\(\Rightarrow A^2>\frac{1}{100}=\frac{1}{10^2}\)

Vậy \(A>\frac{1}{10}\)

\(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{9999}{10000}\)

\(\Rightarrow A>\frac{1}{2}.\frac{2}{3}.\frac{4}{5}...\frac{9998}{9999}\)

\(\Rightarrow A^2>\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}...\frac{9998}{9999}.\frac{9999}{10000}\)

\(\Rightarrow A^2>\frac{1}{10000}=\frac{1}{100^2}\)

\(VayA>\frac{1}{100}=B\)

MỚI LÀM LÚC TỐI,HÊN QUÁ:

\(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\)

\(3A=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\)

\(2A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

\(6A=3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\)

\(4A=3-\left(\frac{101}{3^{99}}-\frac{100}{3^{100}}\right)\)

\(4A=3-\frac{203}{3^{100}}\)

\(A=\frac{3}{4}-\frac{203}{3^{100}\cdot4}< \frac{3}{4}\)

\(\frac{1}{2^1}+\frac{2}{3^2}+\frac{3}{4^3}+...+\frac{99}{100^{99}}<\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}<1\)

Vậy \(\frac{1}{2^1}+\frac{2}{3^2}+\frac{3}{4^3}+...+\frac{99}{100^{99}}<1\)

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

\(\Rightarrow3A=1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{3^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)

\(\Rightarrow A+3A=\left(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\right)+\left(1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{3^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\right)\)

\(\Rightarrow4A=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)<\(B=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\)

\(\Rightarrow3B=3-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{97}}-\frac{1}{3^{98}}\)

\(\Rightarrow B+3B=\left(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\right)+\left(3-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{97}}-\frac{1}{3^{98}}\right)\)

\(\Rightarrow4B=3-\frac{1}{3^{98}}<3\)

\(\Rightarrow B<\frac{3}{4}\Rightarrow4A<\frac{3}{4}\Rightarrow A<\frac{3}{16}\)

\(\RightarrowĐPCM\)

lam ngan hon nua di

\(4S=1+\frac{2}{4}+\frac{3}{4^2}+...+\frac{2019}{4^{2018}}\)

=> \(3S=1+\frac{2}{4}+\frac{3}{4^2}+...+\frac{2019}{2^{2018}}-\frac{1}{4}-\frac{2}{4^2}-\frac{3}{4^3}-...-\frac{2019}{4^{2019}}\)

=>3S=\(1+\frac{1}{4}+\frac{1}{4^2}+..+\frac{1}{2^{2018}}-\frac{2019}{4^{2019}}\)

còn lại tự giải nhé  

Mình cảm ơn bạn.

\(B=\frac{1}{1+3}+\frac{1}{1+3+5}+...+\frac{1}{1+3+...+101}\)

\(B=\frac{1}{4}+\frac{1}{9}+...+\frac{1}{51}\)

\(B=\frac{1}{2\cdot2}+\frac{1}{3\cdot3}+...+\frac{1}{3\cdot17}\)

\(B=\frac{1}{2}-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+...+\frac{1}{3}-\frac{1}{17}\)

\(B=\frac{1}{2}-\frac{1}{17}\)

\(B=\frac{15}{34}\)

TU DO \(=>\frac{15}{34}< \frac{3}{4}\)HOAC \(B< \frac{3}{4}\)

 CHUC BAN HOC TOT :)) 

Ta có: \(1+3=\frac{\left(1+3\right).\left[\left(3-1\right):2+1\right]}{2}=\frac{4.2}{2}=2.2\)

\(1+3+5=\frac{\left(1+5\right).\left[\left(5-1\right):2+1\right]}{2}=\frac{6.3}{2}=3.3\)

                  \(.................\)

\(1+3+5+...+101=\frac{\left(1+101\right).\left[\left(101-1\right):2+1\right]}{2}=\frac{102.5}{2}=51.51\)

\(\Rightarrow B=\frac{1}{2.2}+\frac{1}{3.3}+...+\frac{1}{51.51}\)

\(\Rightarrow B< \frac{1}{2.2}+\frac{1}{2.3}+...+\frac{1}{50.51}=\frac{1}{4}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{50}-\frac{1}{51}\)

\(\Rightarrow B< \left(\frac{1}{4}+\frac{1}{2}\right)-\frac{1}{51}\)

\(\Rightarrow B< \frac{3}{4}-\frac{1}{51}< \frac{3}{4}\)

\(\Rightarrow B>\frac{3}{4}\left(đpcm\right)\)

HÈ RỒI ÍT  NGƯỜI LÀM LẮM

VỚI LẠI LÀ KO BIẾT ĐANG HỌC LỚP 5 LÊN LỚP 6

Bài làm:

Xét: \(\frac{1}{5^2}>\frac{1}{5.6}\) ; \(\frac{1}{6^2}>\frac{1}{6.7}\) ; ... ; \(\frac{1}{100^2}>\frac{1}{100.101}\)

=> \(A>\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{100.101}\)

\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{101}\)

\(=\frac{1}{5}-\frac{1}{101}=\frac{96}{505}>\frac{1}{6}\) (1)

Lại có: \(\frac{1}{5^2}< \frac{1}{4.5}\) ; \(\frac{1}{6^2}< \frac{1}{5.6}\) ; ... ; \(\frac{1}{100^2}< \frac{1}{99.100}\)

=> \(A< \frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)

\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)

\(=\frac{1}{4}-\frac{1}{100}< \frac{1}{4}\) (2)

Từ (1) và (2) => \(\frac{1}{6}< A< \frac{1}{4}\)