Câu hỏi của Nguyễn Thu Ngà

Question

tính $1.C^1_{100}+5.C^2_{100}+9C_{100}^3+....+397C^{100}_{100}$

Nguyễn Việt Lâm · Accepted Answer

$S=1C_{100}^1+\left(4+1\right)C_{100}^2+\left(4.2+1\right)C_{100}^3+...+\left(4.99+1\right)C_{100}^{100}$$=C_{100}^1+C_{100}^2+...+C_{100}^{100}+4\left(1.C_{100}^2+2.C_{100}^3+...+99C_{100}^{100}\right)$$=2^{100}-1+4S_1$Xét khai triển:$\left(1+x\right)^{100}=C_{100}^0+xC_{100}^1+x^2C_{100}^2+...+x^{100}C_{100}^{100}$$\Rightarrow\dfrac{\left(1+x\right)^{100}}{x}=\dfrac{C_{100}^0}{x}+C_{100}^1+xC_{100}^2+...+x^{99}C_{100}^{100}$Đạo hàm 2 vế:$\dfrac{100x\left(1+x\right)^{99}-\left(1+x\right)^{100}}{x^2}=-\dfrac{C_{100}^0}{x^2}+C_{100}^2+2xC_{100}^3+...+99x^{98}C_{100}^{100}$Thay $x=1$$\Rightarrow100.2^{99}-2^{100}=-1+S_1$$\Rightarrow S_1=49.2^{100}+1$$\Rightarrow S=2^{100}-1+4\left(49.2^{100}+1\right)=...$Câu trả lời: $S=1C_{100}^1+\left(4+1\right)C_{100}^2+\left(4.2+1\right)C_{100}^3+...+\left(4.99+1\right)C_{100}^{100}$$=C_{100}^1+C_{100}^2+...+C_{100}^{100}+4\left(1.C_{100}^2+2.C_{100}^3+...+99C_{100}^{100}\right)$$=2^{100}-1+4S_1$Xét khai triển:$\left(1+x\right)^{100}=C_{100}^0+xC_{100}^1+x^2C_{100}^2+...+x^{100}C_{100}^{100}$$\Rightarrow\dfrac{\left(1+x\right)^{100}}{x}=\dfrac{C_{100}^0}{x}+C_{100}^1+xC_{100}^2+...+x^{99}C_{100}^{100}$Đạo hàm 2 vế:$\dfrac{100x\left(1+x\right)^{99}-\left(1+x\right)^{100}}{x^2}=-\dfrac{C_{100}^0}{x^2}+C_{100}^2+2xC_{100}^3+...+99x^{98}C_{100}^{100}$Thay $x=1$$\Rightarrow100.2^{99}-2^{100}=-1+S_1$$\Rightarrow S_1=49.2^{100}+1$$\Rightarrow S=2^{100}-1+4\left(49.2^{100}+1\right)=...$

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