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Nguyễn Huy Tú

tính tổng \(S=2^2.C_{2024}^2-3^2.C_{2024}^3+...+\left(-1\right)^k.k^2.C_{2024}^k+...+2024^2C_{2024}^{2024}\)

Trần Tuấn Hoàng
15 tháng 8 lúc 21:29

\(2\left(S+2023^2.2024-2024^2\right)=2\left(2^2.C^2_{2024}-3^2.C^3_{2024}+...+\left(-1\right)^k.k^2.C^k_{2024}+...+2022^2.C^{2022}_{2024}\right)\)

\(=\left(2^2+2022^2\right).C^2_{2024}-\left(3^2+2021^2\right).C^3_{2024}+...-\left(1011^2+1013^2\right).C^{1011}_{2024}+2.1012^2.C^{1012}_{2024}-\left(1013^2+1011^2\right).C^{1013}_{2024}+...-\left(2021^2+3^2\right).C^{2021}_{2024}+\left(2022^2+2^2\right).C^{2022}_{2024}\)

\(=2024^2\left(C^2_{2024}-C^3_{2024}+....-C^{2021}_{2024}+C^{2022}_{2024}\right)-4.\left(2.2022.C^2_{2024}-3.2021.C^3_{2024}+...-2021.3.C^{2021}_{2024}+2022.2.C^{2022}_{2024}\right)\)

\(=2024^2.\left(\sum\limits^{2022}_{k=2}\left(-1\right)^k.C^k_{2024}\right)-4.\left(\sum\limits^{2022}_{k=2}\left(-1\right)^k.k.\left(2024-k\right).C^k_{2024}\right)\)

Ta có: \(\left(x+1\right)^{2024}=\sum\limits^{2024}_{k=0}C^k_{2024}x^k\left(1\right)\)

\(\Rightarrow\sum\limits^{2024}_{k=0}C^k_{2024}\left(-1\right)^k=0\Rightarrow\sum\limits^{2022}_{k=2}C^k_{2024}\left(-1\right)^k=-C^0_{2024}+C^1_{2024}+C^{2023}_{2024}-C^{2024}_{2024}=4046\)

\(\left(-1\right)^k.k.\left(2024-k\right).C^k_{2024}=\left(-1\right)^k.\dfrac{2024!}{\left(k-1\right)!\left[2022-\left(k-1\right)\right]!}=\left(-1\right)^k.2023.2024.C^{k-1}_{2022}\)

\(\Rightarrow\sum\limits^{2023}_{k=2}\left(-1\right)^k.k.\left(2024-k\right).C^k_{2024}=2023.2024.\left(\sum\limits^{2023}_{k=2}\left(-1\right)^k.C^{k-1}_{2024}\right)\)

Ta có: \(\sum\limits^{2024}_{k=0}C^k_{2024}\left(-1\right)^k=0\Rightarrow\sum\limits^{2024}_{k=0}C^k_{2024}\left(-1\right)^{k+1}=0\Rightarrow\sum\limits^{2022}_{k=2}\left(-1\right)^k.C^{k-1}_{2024}=C^0_{2024}+C^{2022}_{2024}-C^{2023}_{2024}+C^{2024}_{2024}=\dfrac{2023.2024}{2}-2022\)

Vậy: \(2\left(S+2023^2.2024-2024^2\right)=2024^2.4046-4.\left(\dfrac{2023.2024}{2}-2022\right)\)

\(=2023.2024.4046+8088\)

\(\Rightarrow S=\dfrac{2023.2024.4046+8088-2023^2.2024+2024^2}{2}=\dfrac{2023^2.2024+2024^2+8088}{2}\)

 

 


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