Đặt \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+.....+\frac{1}{3^{2012}}\)

\(\Rightarrow3A=1+\frac{1}{3}+\frac{1}{3^2}+......+\frac{1}{3^{2011}}\)

\(\Rightarrow3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+......+\frac{1}{3^{2011}}\right)\)\(-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+.....+\frac{1}{3^{2012}}\right)\)

\(\Rightarrow2A=1-\frac{1}{3^{2012}}\)

\(\Rightarrow A=\frac{1-\frac{1}{3^{2012}}}{2}\)

Vì \(1-\frac{1}{3^{2012}}< 1\Rightarrow A< \frac{1}{2}\)

các bạn giải toán nhanh dùm mình với.mình cần gấp

(1+2+3+4)(1+2+3+4)=(1+2+3+4)²

1.1.1+2.2.2+3.3.3+4.4.4

=1³+2³+3³+4³

=(1+2+3+4)³

=> (1+2+3+4).(1+2+3+4) < 1.1.1+2.2.2+3.3.3+4.4.4

( 1 + 2 + 3 + 4)( 1 + 2 + 3 + 4) = 10 . 10 = 100

1.1.1 + 2.2.2 + 3.3.3 + 4.4.4 = 1 + 8 + 27 + 64  = 100 

VẬy (1 +  2 + 3 + 4)(1 + 2 + 3 + 4) = 1.1.1 + 2.2.2 + 3.3.3 + 4.4.4

Ta có : 

\(T=\frac{2}{2^1}+\frac{3}{2^2}+\frac{4}{2^3}+...+\frac{2016}{2^{2015}}+\frac{2017}{2^{2016}}\) 

\(T=1+\frac{3}{1.2^2}+\frac{4}{2.2^2}+\frac{5}{2^2.2^2}+...+\frac{2016}{2^{2013}.2^2}+\frac{2017}{2^{1014}.2^2}\)

\(=1+\frac{1}{2^2}.\left(3+2+\frac{5}{4}+\frac{6}{8}+...+\frac{2016}{x}+\frac{2017}{x}\right)\)

\(=1+\frac{1}{2^2}.\left(3+2+\frac{5}{2^2}+\frac{6}{2^3}+...+\frac{2016}{2^{2013}}+\frac{2017}{2^{2014}}\right)\)

Đến chỗ này chịu!

Ta có

\(T=1+\frac{3}{1\cdot2^2}+\frac{4}{2\cdot2^2}+...+\frac{2017}{2^2\cdot2^{2014}}\) 

\(T=1+\frac{1}{2^2}\cdot\left(3+2+\frac{5}{2^2}+\frac{6}{2^3}+...+\frac{2016}{2^{2014}}+\frac{2017}{2^{2015}}\right)\)