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Ta có: \(A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2014}}\)
\(2A=2\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2014}}\right)\)
\(2A=2+1+\frac{1}{2}+..+\frac{1}{2^{2013}}\)
\(2A-A=\left(2+1+\frac{1}{2}+..+\frac{1}{2^{2013}}\right)\)\(-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2014}}\right)\)
\(A=2+\left(1+\frac{1}{2}+..+\frac{1}{2^{2013}}\right)\)\(-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2013}}\right)-\frac{1}{2^{2014}}\)
\(A=2-\frac{1}{2^{2014}}\)
Ta có:\(A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2014}}\)
\(\Rightarrow2A=2+1+\frac{1}{2}+...+\frac{1}{2^{2013}}\)
\(\Leftrightarrow2A-A=\left(2+1+\frac{1}{2}+...+\frac{1}{2^{2013}}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2^{2014}}\right)\)
\(=2-\frac{1}{2^{2014}}=\frac{2^{2015}-1}{2^{2014}}\)
Vậy \(A=\frac{2^{2015}-1}{2^{2014}}\)
\(A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2014}}\)
\(\Rightarrow3A=3+1+\frac{1}{3}+...+\frac{1}{3^{2013}}\)
\(\Rightarrow3A-A\)= \(\left(3+1+...+\frac{1}{3^{2013}}\right)-\left(1+\frac{1}{3}+...+\frac{1}{3^{2014}}\right)\)
\(\Rightarrow2A=3-\frac{1}{3^{2014}}\)
\(\Rightarrow A=\frac{3-\frac{1}{3^{2014}}}{2}\)
\(\Rightarrow A=\frac{3}{2}-\frac{\frac{1}{3^{2014}}}{2}< \frac{3}{2}\)
Vậy \(A< \frac{3}{2}\)
Chúc bạn học tốt !!!
a. \(A=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2014}}\)
\(\Rightarrow3A=1+\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{2013}}\)
\(\Rightarrow3A-A=1-\frac{1}{3^{2014}}\)
\(\Rightarrow2A=1-\frac{1}{3^{2014}}\)
\(\Rightarrow A=\left(1-\frac{1}{3^{2014}}\right):2=\frac{1}{2}-\frac{1}{3^{2014}.2}=\frac{3^{2014}-1}{3^{2014}.2}\)
b.\(B=\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{2014}}\)
\(\Rightarrow2B=1+\frac{1}{2^2}+....+\frac{1}{2^{2013}}\)
\(\Rightarrow2B-B=1-\frac{1}{2^{2014}}\)
\(\Rightarrow B=1-\frac{1}{2^{2014}}\)
\(A=\frac{\frac{2017}{1}+\frac{2016}{2}+\frac{2015}{3}+...+\frac{1}{2017}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2018}}\)
\(A=\frac{1+\left(1+\frac{2016}{2}\right)+\left(1+\frac{2015}{3}\right)+...+\left(1+\frac{1}{2017}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2018}}\)
\(A=\frac{\frac{2018}{2018}+\frac{2018}{2}+\frac{2018}{3}+...+\frac{2018}{2017}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2018}}\)
\(A=\frac{2018\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}+\frac{1}{2018}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2018}}\)
\(A=2018\)
Ta có :
\(A=\frac{\frac{2017}{1}+\frac{2016}{2}+\frac{2015}{3}+...+\frac{1}{2017}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}}\)
\(A=\frac{\left(\frac{2017}{1}-1-1-...-1\right)+\left(\frac{2016}{2}+1\right)+\left(\frac{2015}{3}+1\right)+...+\left(\frac{1}{2017}+1\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}}\)
\(A=\frac{\frac{2018}{2018}+\frac{2018}{2}+\frac{2018}{3}+...+\frac{2018}{2017}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}}\)
\(A=\frac{2018\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}}\)
\(A=2018\)
Vậy \(A=2018\)
Chúc bạn học tốt ~
A =\(\frac{1}{1+2}\)+\(\frac{1}{1+2+3}\)+...+\(\frac{1}{1+2+3+4...+2014}\)
\(\Rightarrow A=\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{2029105}\)
\(\Rightarrow2A=2\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{2029105}\right)\)
\(\Rightarrow2A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{4058210}\)
\(\Rightarrow2A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2014.2015}\)
\(\Rightarrow2A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015}\)
\(\Rightarrow2A=\frac{1}{2}-\frac{1}{2015}\)
\(\Rightarrow2A=\frac{2013}{4030}\)
\(\Rightarrow A=\frac{2013}{8060}\)
ngài Kiệt ღ ๖ۣۜLý๖ۣۜ đúng là không ái sánh bằng sự gian xảo này
\(2A=2\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2012}}\right)\)
\(2A=2+1+...+\frac{1}{2^{2011}}\)
\(2A-A=\left(2+1+...+\frac{1}{2^{2011}}\right)-\left(1+\frac{1}{2}+...+\frac{1}{2^{2010}}\right)\)
\(A=2-\frac{1}{2^{2012}}\)
A=\(1\frac{1}{2}.1\frac{1}{3}.1\frac{1}{4}....1\frac{1}{2015}\)
A=\(\frac{3}{2}.\frac{4}{3}.\frac{5}{4}....\frac{2016}{2015}\)
A=\(\frac{3.4.5.....2016}{2.3.4....2015}\)
A=\(\frac{2016}{2}=1008\)
nhanh tay lên các bạn ai trả lời đầu tiên mình cho 5 cái tick
\(A=\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{2014}}\)
\(3A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2013}}\)
\(3A-A=\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2013}}\right)-\left(\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{2014}}\right)\)
\(2A=\frac{1}{3}-\frac{1}{3^{2014}}\)
\(A=\frac{\frac{1}{3}-\frac{1}{3^{2014}}}{2}\)