C>D nhe em ei

2019 nhân 100 thì bằng 201900 > 20203

2020.2021 lớn hơn 100 suy ra b lớn hơn a

\(A=\dfrac{1}{2020}+\dfrac{1}{2020^2}+...+\dfrac{1}{2020^{2021}}\)

\(\Rightarrow2020A=1+\dfrac{1}{2020}+...+\dfrac{1}{2020^{2020}}\)

\(\Rightarrow2020A-A=\left(1+\dfrac{1}{2020}+...+\dfrac{1}{2020^{2020}}\right)-\left(\dfrac{1}{2020}+\dfrac{1}{2020^2}+...+\dfrac{1}{2020^{2021}}\right)\)

\(\Rightarrow2019A=1-\dfrac{1}{2020^{2021}}< 1\Rightarrow A< \dfrac{1}{2019}\)

cảm ơn bạn nhé

bừa nhé: > phải ko?

cảm ơn nhé

A = 1/2^2 + 1/3^2 + 1/4^2 +...+ 1/2020^2

1/2^2 = 1/2.2 < 1/1.2 = 1/1 - 1/2

1/3^2 = 1/3.3 < 1/2.3 = 1/2 - 1/3

1/4^2 = 1/4.4 < 1/3.4 = 1/3 - 1/4

..................................................................

1/2020^2 < 1/2019.2020 = 1/2019 - 1/2020

Cộng vế với vế ta có:

A = 1/2^2 + 1/3^2+..+1/2020^2 = 1/1 - 1/2020

A < 1

1 < 3/2

Vậy A < 1 < 3/2

úi giười ơi hoàng ơi 6a1?

Hello

mn giúp mik vs

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{2019}-\frac{1}{2020}\)

\(=1-\frac{1}{2020}>1\)

Thank you bạn dcv new ^ ^