Ta có:
\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2017}}+\dfrac{1}{3^{2018}}\\ \Rightarrow3A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2016}}+\dfrac{1}{3^{2017}}\)
Lấy 3A trừ A ta được:
\(2A=1-\dfrac{1}{3^{2018}}\\ \Rightarrow A=\dfrac{1-\dfrac{1}{3^{2018}}}{2}\)
Vậy \(A=\dfrac{1-\dfrac{1}{3^{2018}}}{2}\)
Cho: \(C=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{2017}}+\dfrac{1}{3^{2018}}\)
CMR: \(C< \dfrac{1}{2}\)
Cho S=\(\dfrac{1}{5^2}+\dfrac{2}{5^2}+\dfrac{3}{5^3}+\dfrac{4}{5^4}+...+\dfrac{2017}{5^{2017}}+\dfrac{2018}{5^{2018}}\).Chứng minh S<\(\dfrac{1}{3}\)
tìm x biết (\(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+\(\dfrac{1}{4}\)+...+\(\dfrac{1}{2018}\)).x=\(\dfrac{2017}{1}+\dfrac{2016}{2}+\dfrac{2015}{3}+...+\dfrac{2}{2016}+\dfrac{1}{2017}\)
Tìm x :
a) \(3-\dfrac{2}{2x-3}=\dfrac{2}{5}+\dfrac{1}{2x-3}-\dfrac{3}{2}\)
b) \(7.\left(x.1\right)+2x.\left(1-x\right)=0\)
c) \(\dfrac{x+1}{2018}+\dfrac{x+2}{2017}+\dfrac{x+3}{2016}=\dfrac{x+10}{2009}+\dfrac{x+11}{2008}+\dfrac{x+12}{2007}\)
Có 3 ý thui các bạn hãy giúp mk!!!!!!
CM:
\(\dfrac{1}{2^3}+\dfrac{1}{3^3}+\dfrac{1}{4^3}+...+\dfrac{1}{2017^3}< \dfrac{1}{4}\)
Tìm x, biết:
\(\dfrac{3-x}{2016}-1=\dfrac{2-x}{2017}+\dfrac{1-x}{2018}\)
Tính: \(D=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}{\dfrac{2011}{1}+\dfrac{2010}{2}+\dfrac{2009}{3}+...+\dfrac{1}{2011}}\)
Tìm x biết \(\dfrac{x+4}{2015}+\dfrac{x+3}{2016}=\dfrac{x+2}{2017}+\dfrac{x+1}{2018}\)
Tính A=\((1-\dfrac{1}{1+2}).\left(1-\dfrac{1}{1+2+3}\right).\left(1-\dfrac{1}{1+2+3+4}\right)...\left(1-\dfrac{1}{1+2+3+..,+2017}\right)\)
Chúc bn học tốt!
