\(A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2014}}\)

\(3A=3+1+\frac{1}{3}+...+\frac{1}{3^{2013}}\)

\(3A-A=\left(3+1+\frac{1}{3}+...+\frac{1}{3^{2013}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2014}}\right)\)

\(2A=3-\frac{1}{3^{2014}}\)

\(A=\frac{3}{2}-\frac{1}{2.3^{2014}}< \frac{3}{2}\)

\(A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2014}}\)

\(\Rightarrow3A=3+1+\frac{1}{3}+...+\frac{1}{3^{2013}}\)

\(\Rightarrow3A-A=3-\frac{1}{3^{2014}}\)

\(\Rightarrow A=\frac{3}{2}-\frac{1}{3^{2014}.2}< \frac{3}{2}\)

A=1+1/3+1/3^2+...+1/3^2014

3A=3.(1+1/3+1/3^2+...+1/3^2014)

3A=3+1+1/3+....+1/3^2013

Lấy 3A-A ra 2A=3-1/3^2014(nhớ quy tắc phá ngoặc và chuyển dấu nhé)

A=(3-1/3^2014):2=3/2-1/3^2014.2

suy ra A<3/2

Vậy A<3/2

Bài làm của mình có thể có nhiều sai sót mong các bạn sẽ giúp đỡ mình để lần sau bài làm của mình sẽ hoàn thiện hơn

\(A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2014}}\)

\(A=\left(3A-A\right):2\)

\(3A=3+1+\frac{1}{3}+...+\frac{1}{3^{2013}}\)

\(A=\left(3-\frac{1}{3^{2014}}\right):2\)

\(A=\frac{3}{2}-\frac{1}{2.3^{2014}}\)

\(\Rightarrow A<\frac{3}{2}\)

\(A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2014}}\)

\(\Rightarrow3A=3+1+\frac{1}{3}+...+\frac{1}{3^{2013}}\)

\(\Rightarrow3A-A\)=  \(\left(3+1+...+\frac{1}{3^{2013}}\right)-\left(1+\frac{1}{3}+...+\frac{1}{3^{2014}}\right)\)

\(\Rightarrow2A=3-\frac{1}{3^{2014}}\)

\(\Rightarrow A=\frac{3-\frac{1}{3^{2014}}}{2}\)

\(\Rightarrow A=\frac{3}{2}-\frac{\frac{1}{3^{2014}}}{2}< \frac{3}{2}\)

Vậy  \(A< \frac{3}{2}\)

Chúc bạn học tốt !!! 

Ta có:

    A=1+1/3+1/3²+1/3³+...+1/3²⁰¹⁴

=>3A=3+1/3²+1/3³+1/3⁴+...+1/3²⁰¹⁵

=>2A=2+1/3²⁰¹⁵-1/3

=>A=1+2/3²⁰¹⁵-2/3

OK!

Ta có : 

\(A=\frac{1}{2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{18.19.20}\)

\(\Rightarrow A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{18.19.20}\)

\(\Rightarrow A=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{18.19}-\frac{1}{19.20}\right)\)

\(\Rightarrow A=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{19.20}\right)\)

\(\Rightarrow A=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{380}\right)\)

\(\Rightarrow A=\frac{1}{4}-\frac{1}{760}< \frac{1}{4}\)

Vậy \(A< \frac{1}{4}\)

\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{18.19.20}\)

\(\Rightarrow A=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{18.19}-\frac{1}{19.20}\right)\)

\(\Rightarrow A=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{380}\right)=\frac{1}{2}\left(\frac{189}{380}\right)=\frac{189}{760}< \frac{1}{4}\)

Ta có: \(A=\frac{1}{2\times3}+\frac{1}{2\times3\times4}+\frac{1}{3\times4\times5}+....+\frac{1}{18\times19\times20}\)

              \(=\frac{1}{2}\times\left(\frac{1}{1\times2}-\frac{1}{2\times3}+\frac{1}{2\times3}-\frac{1}{3\times4}+...+\frac{1}{18\times19}-\frac{1}{19\times20}\right)\)

               \(=\frac{1}{2}\times\left(\frac{1}{1\times2}-\frac{1}{19\times20}\right)\)

                 \(=\frac{1}{2}\times\frac{1}{1\times2}-\frac{1}{2}\times\frac{1}{19\times20}\)

                   \(=\frac{1}{4}-\frac{1}{2}\times\frac{1}{19\times20}< \frac{1}{4}\)

Vậy A < 1/4

2)Ta có: \(2^{332}< 2^{333}=\left(2^3\right)^{111}=8^{111}\)

              \(3^{223}>3^{222}=\left(3^2\right)^{111}=9^{111}\)

Vì \(8^{111}< 9^{111}\) mà \(2^{332}< 8^{111},3^{223}>9^{111}\) nên suy ra \(2^{332}< 3^{223}\)

Vậy \(2^{332}< 3^{223}\)

1) \(A=\dfrac{10^{2013}+1}{10^{2014}+1}\Rightarrow10A=\dfrac{10^{2014}+10}{10^{2014}+1}=\dfrac{10^{2014}+1}{10^{2014}+1}+\dfrac{9}{10^{2014}+1}=1+\dfrac{9}{10^{2014}+1}\)

\(B=\dfrac{10^{2014}+1}{10^{2015}+1}\Rightarrow10B=\dfrac{10^{2015}+10}{10^{2015}+1}=\dfrac{10^{2015}+1}{10^{2015}+1}+\dfrac{9}{10^{2015}+1}=1+\dfrac{9}{10^{2015}+1}\)Vì: \(10^{2014}+1< 10^{2015}+1\Rightarrow\dfrac{9}{10^{2014}+1}>\dfrac{9}{10^{2015}+1}\Rightarrow1+\dfrac{9}{10^{2014}+1}>1+\dfrac{9}{10^{2015}+1}\)

Nên suy ra \(10A>10B\Rightarrow A>B\)

A=\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{2014.2015.2016}\)

A=\(\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{2014.2015}-\frac{1}{2015.2016}\right)\)

A=\(\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2015.2016}\right)\)

A=\(\frac{1}{4}-\frac{1}{2015.2016.2}\)\(\Rightarrow A<\frac{1}{4}\)