Sửa đề: +2023^2-2024^2

C=(1-2)(1+2)+(3-4)(3+4)+...+(2023-2024)(2023+2024)

=-(1+2+3+4+...+2023+2024)

=-2024*2025/2=-2049300

đề dduk á b

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

Ta có: \(1+\frac12\left(1+2\right)+\frac13\left(1+2+3\right)+\cdots+\frac{1}{2023}\left(1+2+\cdots+2023\right)\)

\(=1+\frac12\cdot\frac{2\cdot3}{2}+\frac13\cdot\frac{3\cdot4}{2}+...+\frac{1}{2023}\cdot\frac{2023\cdot2024}{2}\)

\(=1+\frac32+\frac42+\cdots+\frac{2024}{2}=\frac12\left(2+3+4+\cdots+2024\right)\)

\(=\frac12\left(2024-2+1\right)\cdot\frac{\left(2024+2\right)}{2}=\frac12\cdot2023\cdot\frac{2026}{2}=\frac{2023}{2}\cdot1013\)

=\(\frac{2049299}{2}\)

Ta có: \(B=\frac12+\frac13-\frac14+\frac15-\frac16+\cdots-\frac{1}{2022}+\frac{1}{2023}\)

=>\(B=\frac12+\frac13+\frac14+\frac15+\frac16+\cdots+\frac{1}{2022}+\frac{1}{2023}-2\left(\frac14+\frac16+\cdots+\frac{1}{2022}\right)\)

\(=\frac12+\frac13+\frac14+\frac15+\cdots+\frac{1}{2022}+\frac{1}{2023}-\frac12-\frac13-\cdots-\frac{1}{1011}\)

\(=\frac{1}{1012}+\frac{1}{1013}+\cdots+\frac{1}{2022}+\frac{1}{2023}\)

=C

=>B-C=0

Bằng nhau nha

24^5 K+1

@Bé Bin bạn có biết cách giải không zậy

\(C=\dfrac{2^{2024}-3}{2^{2023}-1}=\dfrac{2.2^{2023}-2-1}{2^{2023}-1}=\dfrac{2\left(2^{2023}-1\right)-1}{2^{2023}-1}=2-\dfrac{1}{2^{2023}-1}\)

\(D=\dfrac{2^{2023}-3}{2^{2022}-1}=\dfrac{2.2^{2022}-2-1}{2^{2022}-1}=\dfrac{2\left(2^{2022}-1\right)-1}{2^{2022}-1}=2-\dfrac{1}{2^{2022}-1}\)

Ta có

\(2^{2023}>2^{2022}\Rightarrow2^{2023}-1>2^{2022}-1\)

\(\Rightarrow\dfrac{1}{2^{2023}-1}< \dfrac{1}{2^{2022}-1}\Rightarrow2-\dfrac{1}{2^{2023}-1}>2-\dfrac{1}{2^{2022}-1}\)

\(\Rightarrow C>D\)