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2A=2*(1+2+22+...+22020)=2+22+...+22021
2A-A=(1+2+22+...+22021)-(1+2+22+...+22020)
A=22021-1<2021
Giải:
A=1+2+22+23+...+22020
2A=2+22+23+24+...+22021
2A-A=(2+22+23+24+...+22021)-(1+2+22+23+...+22020)
A=22021-1
⇒A<22021
Chúc bạn học tốt!
\(B=\frac{2^{2020}+2}{2^{2021}+2}=\frac{2\left(2^{2019}+1\right)}{2\left(2^{2020}+1\right)}=\frac{2^{2019}+1}{2^{2020}+1}\)
vậy A=B=\(\frac{2^{2019}+1}{2^{2020}+1}\)
\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2020}}+\dfrac{1}{2^{2021}}\)
\(\Rightarrow\dfrac{1}{2}A=\dfrac{1}{2}.\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2020}}+\dfrac{1}{2^{2021}}\right)\)\(\Rightarrow\dfrac{1}{2}A=\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{2021}}+\dfrac{1}{2^{2022}}\)
\(\Rightarrow A-\dfrac{1}{2}A=\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2020}}+\dfrac{1}{2^{2021}}\right)-\left(\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{2021}}+\dfrac{1}{2^{2022}}\right)\)\(\Rightarrow\dfrac{1}{2}A=\dfrac{1}{2}-\dfrac{1}{2^{2022}}\)
\(\Rightarrow\dfrac{1}{2}A=\dfrac{2^{2021}-1}{2^{2022}}\)
\(\Rightarrow A=\dfrac{2^{2021}-1}{2^{2023}}.2=\dfrac{2^{2021}-1}{2^{2021}}\)
Vậy \(A=\dfrac{2^{2021}-1}{2^{2021}}\)
A = 1 + 2 + 22 + ... + 22021
2A = 2 + 4 + 23 + ... 22022
A = 22022 - 1
A=1/2+1/22+1/23+...+1/22020+1/22021 > B=1/3+1/4+1/5+13/60
\(A=1+2+2^2+...+2^{2020}+2^{2021}\\ \Rightarrow2A=2+2^2+2^3+...+2^{2021}+2^{2022}\\ \Rightarrow2A-A=A=2^{2022}-1\)
Vậy \(A\) và \(B\) là 2 số tự nhiên liên tiếp.
siêu vậy
a) Đặt A = 2.11 + 2.13 + ... + 2.29
= 2.(11 + 13 + 15 + ... + 29)
Đặt B = 11 + 13 + 15 + ... + 29
Số số hạng của B:
(29 - 11) : 2 + 1 = 10 (số)
A = 2.(29 + 11) . 10 : 2
= 40.10
= 400
b) (2²⁰²² + 2²⁰²¹- 2²⁰²⁰) : (2²⁰¹⁹ . 2)
= 2²⁰²⁰.(2² + 2 - 1) : 2²⁰²⁰
= 4 + 2 - 1
= 5
\(B=\frac{2^{2020}+2}{2^{2021}+2}\)
\(=\frac{2\left(2^{2019}+1\right)}{2\left(2^{2020}+1\right)}\)
\(=\frac{2^{2019}+1}{2^{2020}+1}=A\)
Vậy \(A=B\)
P/s: Bài này mk thường thấy dạng như phía dưới, bn đọc tham khảo
\(B=\frac{2^{2020}+1}{2^{2021}+1}< \frac{2^{2020}+1+1}{2^{2021}+1+1}=\frac{2^{2020}+2}{2^{2021}+2}=\frac{2^{2019}+1}{2^{2020}+1}=A\)
Vậy \(A>B\)