\(\dfrac{C_n^k}{\left(k+1\right)\left(k+2\right)}=\dfrac{n!}{\left(k+1\right)\left(k+2\right).k!\left(n-k\right)!}=\dfrac{1}{\left(n+1\right)\left(n+2\right)}.\dfrac{\left(n+2\right)!}{\left(n+2-\left(k+2\right)\right)!\left(k+2\right)!}\)

\(=\dfrac{1}{\left(n+1\right)\left(n+2\right)}.C_{n+2}^{k+2}\)

Đặt tổng trên là A

\(\Rightarrow A=\dfrac{-1.C_{2024}^3}{2023.2024}+\dfrac{2.C_{2024}^4}{2023.2024}+\dfrac{-3.C_{2024}^5}{2023.2024}+...+\dfrac{2022.C_{2024}^{2024}}{2023.2024}\)

\(=\dfrac{1}{2023.2024}\left(-1.C_{2024}^3+2.C_{2024}^4+...+2022.C_{2024}^{2024}\right)=\dfrac{1}{2023.2024}.B\)

Xét \(C=-2.\left(-C_{2024}^3+C_{2024}^4-C_{2024}^5+...+C_{2024}^{2024}\right)\)

\(\Rightarrow B-C=-3C_{2024}^3+4C_{2024}^4-5C_{2024}^5+...+2024.C_{2024}^{2024}\)

Ta có:

\(k.C_n^k=\dfrac{n!.k}{\left(n-k\right)!.k!}=n.\dfrac{\left(n-1\right)!}{\left(\left(n-1\right)-\left(k-1\right)\right)!.\left(k-1\right)!}=n.C_{n-1}^{k-1}\)

\(\Rightarrow B-C=-2024.C_{2023}^2+2024C_{2023}^3+...+2024.C_{2023}^{2023}\)

\(=-2024\left(C_{2023}^2-C_{2023}^3+...-C_{2023}^{2023}\right)\)

Xét khai triển:

\(\left(1-x\right)^k=C_k^0-xC_k^1+x^2C_k^2+...+\left(-1\right)^kx^k.C_k^k\)

Thay \(k=2024\); \(x=1\)

\(\Rightarrow0=C_{2024}^0-C_{2024}^1+C_{2024}^2-C_{2024}^3+...+C_{2024}^{2024}\)

\(\Rightarrow-C_{2024}^3+...+C_{2024}^{2024}=C_{2024}^1-C_{2024}^2-1\)

\(\Rightarrow C=-2\left(C_{2024}^1-C_{2024}^2-1\right)=-2\left(2023-C_{2024}^2\right)\)

Thay \(k=2023;x=1\)

\(\Rightarrow0=C_{2023}^0-C_{2023}^1+C_{2023}^2+...-C_{2023}^{2023}\)

\(\Rightarrow C_{2023}^2-C_{2023}^3+...-C_{2023}^{2023}=C_{2023}^1-1=2022\)

\(\Rightarrow B-C=-2024.2022\)

\(\Rightarrow B=C-2022.2024=-2\left(2023-C_{2024}^2\right)-2022.2024\)

\(=-2.2023+2023.2024-2022.2024\)

\(=-2022\)

\(\Rightarrow A=\dfrac{-2022}{2023.2024}\)

\(\Leftrightarrow1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{2021}{2022}\)

\(\Leftrightarrow1-\dfrac{1}{x+1}=\dfrac{2021}{2022}\)

\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{2022}\)

=>x+1=2022

hay x=2021

3a-b=1/2(a+b)

=>6a-2b=a+b

=>5a=3b

=>a/3=b/5=k

=>a=3k; b=5k

\(A=\dfrac{a^{2022}+3^{2022}}{b^{2022}+5^{2022}}\)

\(=\dfrac{3^{2022}\left(k^{2022}+1\right)}{5^{2022}\left(k^{2022}+1\right)}=\left(\dfrac{3}{5}\right)^{2022}\)

A>B

A>B

nếu cần mik chụp cho bạn 1 đề tương tự

504

Trước hết ta phải chứng minh \(\dfrac{a}{b}< \dfrac{a+1}{b+1}\) (a, b ϵ N; a < b).

Thật vậy, \(\dfrac{a}{b}=\dfrac{a\left(b+1\right)}{b\left(b+1\right)}=\dfrac{a+ab}{b^2+b}\) và \(\dfrac{a+1}{b+1}=\dfrac{\left(a+1\right)b}{\left(b+1\right)b}=\dfrac{ab+b}{b^2+b}\).

Mà theo giả thuyết là a < b nên \(\dfrac{a+ab}{b^2+b}< \dfrac{ab+b}{b^2+b}\), suy ra \(\dfrac{a}{b}< \dfrac{a+1}{b+1}\) (a, b ϵ N; a < b).

Từ đây ta có:

\(B=\dfrac{2022^{2022}+1}{2022^{2023}+1}=\dfrac{2022^{2023}+2022}{2022^{2024}+2022}=\dfrac{2022^{2023}+2021+1}{2022^{2024}+2021+1}\)

Đặt \(A_1=\dfrac{2022^{2023}+2}{2022^{2024}+2}=\dfrac{2022^{2023}+1+1}{2022^{2024}+1+1}\), rõ ràng \(A_1>A\).

Đặt \(A_2=\dfrac{2022^{2023}+3}{2022^{2024}+3}=\dfrac{2022^{2023}+2+1}{2022^{2024}+2+1}\), rõ ràng \(A_2>A_1\).

...

Đặt \(A_{2020}=\dfrac{2022^{2023}+2021}{2022^{2024}+2021}=\dfrac{2022^{2023}+2020+1}{2022^{2024}+2020+1}\), rõ ràng \(A_{2020}>A_{2019}\) và \(B>A_{2020}\).

Suy ra \(B>A_{2020}>A_{2019}>...>A_2>A_1>A\). Vậy A < B.

Ta có A = \(\dfrac{2022^{2023}}{2022^{2024}}=\dfrac{1}{2022}\) ; B = \(\dfrac{2022^{2022}}{2022^{2023}}=\dfrac{1}{2022}\)

Mà \(\dfrac{1}{2022}=\dfrac{1}{2022}\)

Vậy A = B

 \(=\dfrac{2021}{2022}\left(\dfrac{6}{17}-\dfrac{23}{17}\right)+\dfrac{2021}{2022}=\dfrac{-2021}{2022}+\dfrac{2021}{2022}=0\)