A = \(\dfrac{1}{1+2+3}\)+\(\dfrac{1}{1+2+3+4}\)+...+ \(\dfrac{1}{1+2+...+2004}\)+ \(\dfrac{2}{2025}\)

A = \(\dfrac{1}{\left(1+3\right).3:2}\)+\(\dfrac{1}{\left(4+1\right).4:2}\)+...+ \(\dfrac{1}{\left(2024+1\right).2024:2}\)+\(\dfrac{2}{2025}\)

A = \(\dfrac{2}{3.4}\)+\(\dfrac{2}{4.5}\)+...+\(\dfrac{2}{2024.2025}\)+ \(\dfrac{2}{2025}\)

A = 2.(\(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\)+...+ \(\dfrac{1}{2024.2025}\)) + \(\dfrac{2}{2025}\)

A = 2.(\(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\)+...+ \(\dfrac{1}{2024}\) - \(\dfrac{1}{2025}\)) + \(\dfrac{2}{2025}\)

A = 2.(\(\dfrac{1}{3}\) - \(\dfrac{1}{2025}\)) + \(\dfrac{2}{2025}\)

A = \(\dfrac{2}{3}\) - \(\dfrac{2}{2025}\) + \(\dfrac{2}{2025}\)

A  = \(\dfrac{2}{3}\) 

Cc

Đề có phải là:

\(\dfrac{x+1}{2024}+\dfrac{x+2}{2025}+\dfrac{x+3}{2026}+\dfrac{x+4}{2027}=4\text{ ?}\)

\(\Rightarrow\text{ }\dfrac{x+1}{2024}+\dfrac{x+2}{2025}+\dfrac{x+3}{2026}+\dfrac{x+4}{2027}-4=0\)

\(\Rightarrow\text{ }\dfrac{x+1}{2024}+\dfrac{x+2}{2025}+\dfrac{x+3}{2026}+\dfrac{x+4}{2027}-1-1-1-1=0\)

\(\Rightarrow\left(\dfrac{x+1}{2024}-1\right)+\left(\dfrac{x+2}{2025}-1\right)+\left(\dfrac{x+3}{2026}-1\right)+\left(\dfrac{x+4}{2027}-1\right)=0\)

\(\Rightarrow\left(\dfrac{x+1-2024}{2024}\right)+\left(\dfrac{x+2-2025}{2025}\right)+\left(\dfrac{x+3-2026}{2026}\right)+\left(\dfrac{x+4-2027}{2027}\right)=0\)

\(\Rightarrow\dfrac{x-2023}{2024}+\dfrac{x-2023}{2025}+\dfrac{x-2023}{2026}+\dfrac{x-2023}{2027}=0\)

\(\Rightarrow\left(x-2023\right)\left(\dfrac{1}{2024}+\dfrac{1}{2025}+\dfrac{1}{2026}+\dfrac{1}{2027}\right)=0\)

Mà \(\dfrac{1}{2024}+\dfrac{1}{2025}+\dfrac{1}{2026}+\dfrac{1}{2027}\ne0\)

\(\Rightarrow x-2023=0\)

\(\Rightarrow x=0+2023\)

\(\Rightarrow x=2023\)

Vậy, \(x=2023.\)

\(1:\dfrac{2}{3}:\dfrac{3}{4}:\dfrac{4}{5}:...:\dfrac{2024}{2025}\)

= \(1\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot...\cdot\dfrac{2025}{2024}=\dfrac{2025}{2}\)

\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2024}}\\ =>2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2023}}\\ =>2A-A=A=1-\dfrac{1}{2^{2024}}=\dfrac{2^{2024}-1}{2^{2024}}\)

e cảm ơn

Giúp mình vs ạ

A = 1 - 2 - 3 + 4 + 5 - 6 - 7 + 8 + 9 - 10 - 11 + ... - 2023 + 2024 + 2025

Xét dãy số: 1; 2; 3; 4;..; 2025 là dãy số cách đều với khoảng cách là:

                   2  - 1  = 1

Số số hạng của dãy số trên là: ( 2025 - 1) : 1  + 1 = 2025

                  Vì 2025 : 4 = 506 dư 1 

Nhóm 4 số hạng liên tiếp của A vào nhau thì được A là tổng của 506 nhóm và 2025 khi đó

A =(1-2-3+4)+(5 - 6 - 7 + 8) +...+(2021-2022-2023+2024) + 2025

A = 0 + 0 +...+ 0 + 2025

A = 2025

a: Ta có: \(A=2+2^2+2^3+\cdots+2^{2025}\)

=>\(2A=2^2+2^3+2^4+\cdots+2^{2026}\)

=>\(2A-A=2^2+2^3+2^4+\cdots+2^{2026}-2-2^2-\cdots-2^{2025}\)

=>\(A=2^{2026}-2\)

b:Sửa đề: \(B=1+5+5^2+\cdots+5^{150}\)

=>\(5B=5+5^2+5^3+\cdots+5^{151}\)

=>\(5B-B=5+5^2+5^3+\cdots+5^{151}-1-5-5^2-\cdots-5^{150}\)

=>\(4B=5^{151}-1\)

=>\(B=\frac{5^{151}-1}{4}\)

c: Ta có: \(C=3+3^2+3^3+\ldots+3^{1000}\)

=>\(3C=3^2+3^3+3^4+\cdots+3^{1001}\)

=>\(3C-C=3^2+3^3+\cdots+3^{1001}-3-3^2-\cdots-3^{1000}\)

=>\(2C=3^{1001}-3\)

=>\(C=\frac{3^{1001}-3}{2}\)

B = 1 + 2\(^2\) + 2\(^4\) + ... + 2\(^{2024}\) + 2\(^{2026}\)

2\(^2\) B = 2\(^2\) + 2\(^4\) + ...+ \(2^{2026}\) + 2\(^{2028}\)

4B - B = 2\(^2\) + 2\(^4\) + ...+ \(2^{2026}\) + 2\(^{2028}\) - 1 - 2\(^2\) - 2\(^4\) - ... - 2\(^{2024}\) - 2\(^{2026}\)

3B = (2\(^2\) - 2\(^2\)) + (2\(^4\) - 2\(^4\)) +...+ (2\(^{2026}\) - \(2^{2026}\)) + (2\(^{2028}\) - 1)

3B = 0 + 0 +... + 0 + 2\(^{2028}\) - 1

3B = 2\(^{2028}\) - 1

B = \(\frac{2^{2028}-1}{3}\)

Ta có: \(B=1+2^2+2^4+\cdots+2^{2024}+2^{2026}\)

=>\(4B=2^2+2^4+2^6+\ldots+2^{2026}+2^{2028}\)

=>\(4B-B=2^2+2^4+2^6+\cdots+2^{2026}+2^{2028}-1-2^2-\cdots-2^{2026}\)

=>\(3B=2^{2028}-1\)

=>\(B=\frac{2^{2028}-1}{3}\)

Đặt \(A=1+2+2^2+2^3+\cdots+2^{2023}\)

=>\(2A=2+2^2+2^3+2^4+\cdots+2^{2024}\)

=>\(2A-A=2+2^2+2^3+2^4+\cdots+2^{2024}-1-2-\cdots-2^{2023}\)

=>\(A=2^{2024}-1\)

Ta có: \(2^{2024}-\left(1+2+2^2+2^3+\cdots+2^{2023}\right)\)

\(=2^{2024}-\left(2^{2024}-1\right)=1\)

2^ 2024 -( 1 + 2 + 2 ^ 2 + 2 ^ 3 +...+2^ 2023 )

2^2024 -(2^2024-1)

1

Nhưng dài lắm chắc lớp 6 ko hiểu đc nên tui chỉ viết ngắn gọn

BẠN THÔNG CẢM

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