\(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2020}\)
\(=\dfrac{1}{1+2+3+...+2020}\)
\(=\dfrac{1}{\left(2020-1\right)+1}\)
\(=\dfrac{1}{2020}\)
1/2+1/3+2/3+1/4+2/4+3/4+......+1/2020+2=2020+3/2020+.....+2019/2020
1/2+1/3+2/3+1/4+2/4+3/4+......+1/2020+2/2020+3/2020+.....+2019/2020
Tính:
[2020(1+1/2+1/3+...+1/2020)]/1/2+2/3+3/4+...+2019/2020
Giúp với, mình cần gấp!
a,1/2+1/3+1/4+...+1/2020
b,1/2+1/2^2+...+1/2^109
c,1/1*2+1/2*3+...1/2020*2021
Câu 24: Cho biểu thức: A=1/2+1/3+1/4+.........+1/2021+1/2022 Và B=2021/1+2020/2+2019/3+.........+3/2019+2020+1/2021
cho A=1 1/2 . 1 1/3 . 1 1/4 ....... 1 1/2020
B=(1-1/2) . (1-1/3) . (1- 1/4) ....... (1- 1/2020)
tính tỉ số của A và B
Chứng minh rằng1/2008+1/2009+1/2010+.........+1/2020=1-1/2+1/3-1/4+.......+1/2019-1/2020
Cho A = 2^2018 / 2^2018 + 3^2019 +3^2019/ 3^2019 + 5 ^ 2020 +5^2020 / 5^2020 + 2^2018
B=1/1*2+1/3*4+1/5*6+. . . +1/2019*2020
So sánh A và B
Giúp mk, mk kick cho
tìm A = 2/1+2 + 5/1+2+3 + 9/1+2+3+4 + ..... + 2041210/1+2+3+4 + ...+ 2020
