1+1/2.(1+2)+1/3.(1+2+3)+1/4.(1+2+3+4)+...+1/2023.(1+2+3+...+2023)

=1+1/2.(1+2).2/2+1/3.(1+3).3/2+1/4.(1+4).4/2+...+1/2023.(1+2+3+...+2023).2023/2

=2/2+3/2+4/2+...+2023/2

=2+3+4+...+2023/2

=2025.2022/2/2                 

=1023637,5                        

tham khảo thôi nha

1:

a: =23/27-11/17+4/27+28/17

=23/27+4/27+28/17-11/17

=1+1=2

b: \(=\dfrac{2}{3}\cdot\left(\dfrac{7}{9}+\dfrac{2}{9}\right)-\dfrac{2}{9}\)

=2/3-2/9

=6/9-2/9

=4/9

c: \(=\dfrac{11}{5}\cdot\dfrac{7}{3}-\dfrac{1}{3}\cdot\dfrac{11}{5}\)

=11/5(7/3-1/3)

=11/5*2

=22/5

d: \(=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{2024}{2023}=\dfrac{2024}{2}=1012\)

e: \(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{2022}{2023}=\dfrac{1}{2023}\)

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 

Lời giải:
Gọi tổng trên là $A$
$A=\frac{1}{\frac{3.4}{2}}+\frac{1}{\frac{4.5}{2}}+....+\frac{1}{\frac{2023.2024}{2}}$

$=\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{2023.2024}$

$=2(\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{2024-2023}{2023.2024})$

$=2(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{2023}-\frac{1}{2024})$

$=2(\frac{1}{3}-\frac{1}{2024})=\frac{2021}{3036}$

A=23.4​1​+24.5​1​+....+22023.2024​1​

=23.4+24.5+...+22023.2024=3.42​+4.52​+...+2023.20242​

=2(4−33.4+5−44.5+...+2024−20232023.2024)=2(3.44−3​+4.55−4​+...+2023.20242024−2023​)

=2(13−14+14−15+....+12023−12024)=2(31​−41​+41​−51​+....+20231​−20241​)

=2(13−12024)=20213036=2(31​−20241​)=30362021​
 

\(=\dfrac{1-2^2}{2^2}\cdot\dfrac{1-3^2}{3^2}\cdot...\cdot\dfrac{1-2023^2}{2023^2}\)

\(=\dfrac{2^2-1}{2^2}\cdot\dfrac{3^2-1}{3^2}\cdot...\cdot\dfrac{2023^2-1}{2023^2}\)

\(=\dfrac{1}{2}\cdot\dfrac{3}{2}\cdot\dfrac{2}{3}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{2022}{2023}\cdot\dfrac{2024}{2023}\)

\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{2022}{2023}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{2024}{2023}\)

\(=\dfrac{1}{2023}\cdot\dfrac{2024}{2}=\dfrac{1012}{2023}\)

=1012/2023