Tính a + b + c - d.

Câu 37: Ta có: \[(1+x)^{2024}=C^0_{2024}+C^1_{2024}x+C^2_{2024}x^2+C^3_{2024}x^3+C^4_{2024}x^4+...+C^{2024}_{2024}x^{2024}\] Do đó: \[x(1+x)^{2024}=C^0_{2024}x+C^1_{2024}x^2+C^2_{2024}x^3+C^3_{2024}x^4+C^4_{2024}x^5+...+C^{2024}_{2024}x^{2025}\] Lấy đạo hàm hai vế ta được: \[(1+x)^{2024}+2024x(1+x)^{2023}=C^0_{2024}+2C^2_{2024}x+3C^3_{2024}x^2+4C^4_{2024}x^3+...+2025C^{2024}_{2024}x^{2024}\] Thay \(x\) bởi \(-x^2\) ta được: \[(1-x^2)^{2024}-2024x^2(1-x^2)^{2023}=C^0_{2024}-2C^2_{2024}x^2+3C^3_{2024}x^4-4C^4_{2024}x^6+...-2025C^{2024}_{2024}x^{4048}\) Chia cả hai vế cho \(x\) rồi lấy tích phân từ \(0\) đến \(1\) ta được: \[\int_0^1\frac{(1-x^2)^{2024}-2024x^2(1-x^2)^{2023}}xdx=\int_0^1[C^0_{2024}-2C^2_{2024}x^2+3C^3_{2024}x^4-4C^4_{2024}x^6+...-2025C^{2024}_{2024}x^{4048}]dx\] \(\Rightarrow \int_0^1\frac{(1-x^2)^{2024}-2024x^2(1-x^2)^{2023}}xdx=C^0_{2024}+\frac12C^2_{2024}+\frac13C^4_{2024}+\frac14C^6_{2024}+...+\frac1{1013}C^{2024}_{2024}\) \(\Rightarrow S=\int_0^1\frac{(1-x^2)^{2024}-2024x^2(1-x^2)^{2023}}xdx\) \(\Rightarrow S=\int_0^1\frac{(1-x^2)^{2023}(1-2024x^2)}xdx\) Đặt \(t=1-x^2\Rightarrow dt=-2xdx\) Đổi cận \(x=0\Rightarrow t=1;x=1\Rightarrow t=0\) \(\Rightarrow S=\int_1^0\frac{t^{2023}(1-2024(1-t))}{1-t}(-\frac{dt}{2})\) \(\Rightarrow S=\frac12\int_0^1\frac{t^{2023}(2023t-2023)}{1-t}dt\) \(\Rightarrow S=\frac{2023}{2}\int_0^1\frac{t^{2024}-t^{2023}}{1-t}dt\) \(\Rightarrow S=\frac{2023}{2}\int_0^1\frac{t^{2023}(t-1)}{1-t}dt\) \(\Rightarrow S=\frac{2023}{2}\int_0^1t^{2023}dt\) \(\Rightarrow S=\frac{2023}{2}.\frac1{2024}\) \(\Rightarrow S=\frac{2023}{4048}\) \(\Rightarrow S=2.[\frac{2023^{2024}}{4048}-\frac{2023^{2025}-1}{4048.2023}]\) Vậy \(a+b+c-d=2023+4048+2023-2023=4048\) Đáp án: 4048