Các câu hỏi tương tự
\(A=\frac{\frac{2017}{1}+\frac{2016}{2}+\frac{2015}{3}+...+\frac{1}{2017}+2017}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}\)
Chứng minh A là một số nguyên dương :
A = \(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}+\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}\right)^2+\)\(\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}\right)^2+...+\left(\frac{1}{2017}\right)^2\)
tính :
\(A=\frac{\frac{2017}{1}+\frac{2017}{2}+\frac{2017}{3}+...+\frac{2017}{2018}}{\frac{2017}{1}+\frac{2016}{2}+\frac{2015}{3}+...+\frac{1}{2017}}\)
giúp mình đc ko mọi ng ???
ai nhanh nhất mình k cho nha.
Chứng minh
\(A=1-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{2017}}>\frac{1}{2^{2018}}\)
Giúp mk nha thanks nhiều
Tính nhanh:
A= 1/2+1/2^2+1/2^3+....+1/2^100
B=3^2/2x4+3^2/4x6+3^2/6x8+....+3^2/198x200
C=\(\frac{\frac{2017}{1}+\frac{2017}{2}+\frac{2017}{3}+...+\frac{2017}{2017}}{\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+...+\frac{1}{2016}}\)
D=1x2+2x3+3x4+4x5+...+48x49
E=\(^{1^2+2^2+3^2+...+48^2}\)
F=1x49+2x48+3x47+...+48x2+49x1
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2017}{2019}\)
Giúp mk nha
Tìm x biết
a) \(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\right).x=\frac{2018}{1}+\frac{2017}{2}+\frac{2016}{3}+...+\frac{2}{2017}+\frac{1}{2018}\)
b) \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2017}{2019}\)
tính
A=\(\frac{\frac{2017}{2}+\frac{2017}{3}+\frac{2017}{4}+...+\frac{2017}{2018}}{\frac{2017}{1}+\frac{2016}{2}+...+\frac{1}{2017}}\)
\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}\right).x+2017=\frac{2018}{1}+\frac{2018}{2}+\frac{2020}{3}\). Tìm x