Số hạng của dãy số trên là : \(\left(2026-1\right):1+1\text{=}2026\)

Ta xét với cặp : 1-2 ; 3-4 ; ......... ; 2025-2026=-1

Tổng của dãy số trên là : \(\dfrac{\left(1-2\right).2026}{2}\text{=}-1013\)

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, \(x=2023.\)

ta có : \(\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+\dfrac{1}{\sqrt{4}+\sqrt{5}}+...+\dfrac{1}{\sqrt{2025}+\sqrt{2026}}\)

\(=\dfrac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{3}-\sqrt{2}\right)}+\dfrac{\sqrt{4}-\sqrt{3}}{\left(\sqrt{3}+\sqrt{4}\right)\left(\sqrt{4}-\sqrt{3}\right)}+...+\dfrac{\left(\sqrt{2026}-\sqrt{2025}\right)}{\left(\sqrt{2026}+\sqrt{2025}\right)\left(\sqrt{2026}-\sqrt{2025}\right)}\)

\(=\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+\sqrt{5}-\sqrt{4}+...+\sqrt{2026}-\sqrt{2025}\)

\(=-\sqrt{2}+\sqrt{2026}\)

Hai bài trên áp dụng công thức với khoảng cách là 2.

Ta có:

\(D=1+2^1+2^2+2^3+.....+2^{150}\)

\(\Rightarrow2D-D=\left(2+2^2+2^3+2^4+.....+2^{151}\right)-\left(1+2+2^2+2^3+....+2^{150}\right)\)

\(\Rightarrow D=2^{151}-1\)

\(E=1+4^1+4^2+....+4^{400}\)

\(\Rightarrow4E-E=\left(4+4^2+4^3+....+4^{401}\right)-\left(1+4^1+4^2+....+4^{400}\right)\)

\(\Rightarrow E\left(4-1\right)=4^{401}-1\Leftrightarrow E=\frac{4^{401}-1}{4-1}\)

Các câu còn lại làm tương tự

Ta có 2 trường hợp:

Khi n > 2026:
2^m + 2025 = n - 2026 + n - 2026
2^m + 4051 = 2n - 4052
2^m + 4052 = 2n

Khi n < 2026:
2^m + 2025 = 2026 - n + 2026 - n
2^m + 4051 = 4052 - 2n
2^m + 2n = 4052 - 4051
2^m + 2n = 1

1x5=

1) Ta thấy:

\(4=1+3=1+\sqrt{9}\)

\(1+2\sqrt{2}=1+\sqrt{2^2\cdot2}=1+\sqrt{8}\)

Mà: \(\sqrt{8}< \sqrt{9}\)

\(\Rightarrow1+\sqrt{8}< 1+\sqrt{9}\)

\(\Rightarrow\dfrac{1}{1+\sqrt{8}}>\dfrac{1}{1+\sqrt{9}}\)

\(\Rightarrow\dfrac{1}{1+2\sqrt{2}}>\dfrac{1}{4}\)

2) Ta thấy:

\(2018< 2024\)

\(\Rightarrow\sqrt{2018}< \sqrt{2024}\) (1)

\(2025< 2026\)

\(\Rightarrow\sqrt{2025}< \sqrt{2026}\) (2)

Từ (1) và (2) ta có:

\(\sqrt{2018}+\sqrt{2025}< \sqrt{2024}+\sqrt{2026}\)

2025 x 2026 - 2026 - 1024 x 2 x 2013

= 2026 x (2025 - 1) - 2048 x 2013

= 2026 x 2024 - 2048 x 2013

= (2048 - 22) x 2024 - 2048 x 2013

= 2048 x 2024 - 22 x 2024 - 2048 x 2013

= 2048 x (2024 - 2013) - 22 x 2024

= 2048 x 11 - 22 x 2024

= 2048 x 11 - 2 x 11 x 2024

= 11 x (2048 - 2 x 2024)

= 11 x (2048 - 4048)

= 11 x (-2000)

= -22000

= 2026 x (2025 - 1) - 2048 x 2013

= 2026 * 2024 - 2048 * 2013

= 4100624 - 4122624

= -22000

Oke

Cảm ơn bn nhé