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2020/2021<1
2021/2022<1
2022/2023<1
2023/2020=1+1/2020+1/2020+1/2020>1+1/2021+1/2022+1/2023
=>B>2020/2021+2021/2022+2022/2023+1/2021+1/2022+1/2023+1=4
B = \(\dfrac{1}{2002}\) + \(\dfrac{2}{2021}\) + \(\dfrac{3}{2020}\)+...+ \(\dfrac{2021}{2}\) + \(\dfrac{2022}{1}\)
B = \(\dfrac{1}{2002}\) + \(\dfrac{2}{2021}\) + \(\dfrac{3}{2020}\)+...+ \(\dfrac{2021}{2}\) + 2022
B = 1 + ( 1 + \(\dfrac{1}{2022}\)) + ( 1 + \(\dfrac{2}{2021}\)) + \(\left(1+\dfrac{3}{2020}\right)\)+ ... + \(\left(1+\dfrac{2021}{2}\right)\)
B = \(\dfrac{2023}{2023}\) + \(\dfrac{2023}{2022}\) + \(\dfrac{2023}{2021}\) + \(\dfrac{2023}{2020}\) + ...+ \(\dfrac{2023}{2}\)
B = 2023 \(\times\) ( \(\dfrac{1}{2023}\) + \(\dfrac{1}{2022}\) + \(\dfrac{1}{2021}\) + \(\dfrac{1}{2020}\)+ ... + \(\dfrac{1}{2}\))
Vậy B > C
Đặt B=2023−2022+2021−2020+...+3−2+1�=2023-2022+2021-2020+...+3-2+1
B=(2023−2022)+(2021−2020)+...+(3−2)+1�=(2023-2022)+(2021-2020)+...+(3-2)+1
Đặt A=(2023−2022)+(2021−2020)+...+(3−2)�=(2023-2022)+(2021-2020)+...+(3-2)
Biểu thức A� có số số hạng là:
(2023−2):1+1=2022(2023-2):1+1=2022 (số hạng)
Số nhóm được lập là:
2022:2=10112022:2=1011 (nhóm)
A=1+1+...+1�=1+1+...+1 [10111011 số hạng]
A=1×1011=1011�=1×1011=1011
⇒B=1011+1=1012⇒�=1011+1=1012
Vậy B=1012
Sửa đề: 1-2-3+4+5-6-7+8+...-2018-2019+2020+2021-2022-2023
=(1-2-3+4)+(5-6-7+8)+...+(2017-2018-2019+2020)+(2021-2022-2023)
=0+0+...+0+(-1-2023)
=-2024
2011+2012+2013+2014+2015+2016+2017+2018+2019+2020+2021+ 2022+2023 =(2011+2023)+(2013+2022)+...+(2016+2018)+2017 =4034+4034+4034+4034+4034+4034+2017 =4034x6+2017=26221
2011+2012+2013+2014+2015+2016+2017+2018+2019+2020+2021+2022+2023
=(2011+2023)+(2013+2022)+...+(2016+2018)+2017 =4034+4034+4034+4034+4034+4034+2017 =4034x6+2017=26221