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Xét khai triển:
\(\left(1+x\right)^n=C_n^0+xC_n^1+x^2C_n^2+...+x^nC_n^n\)
Đạo hàm 2 vế:
\(n\left(1+x\right)^{n-1}=C_n^1+2xC_n^2+...+nx^{n-1}C_n^n\)
Tiếp tục đạo hàm 2 vế:
\(\left(n-1\right)n\left(1+x\right)^{n-2}=2C_n^2+2.3xC_n^3+...+\left(n-1\right)nx^{n-2}C_n^n\)
Thay \(x=1\)
\(\Rightarrow\left(n-1\right)n.2^{n-2}=1.2C_n^2+2.3C_n^3+...+\left(n-1\right)nC_n^n\)
\(\Rightarrow\left(n-1\right)n.2^{n-2}+n=C_n^1+1.2C_n^2+...+\left(n-1\right)n.C_n^n\)
\(\Rightarrow S=\left(n-1\right)n.2^{n-2}+n\)
Xét khai triển:
\(\left(1+x\right)^n=C_n^0+xC_n^1+x^2C_n^2+...+x^nC_n^n\)
Đạo hàm 2 vế:
\(n\left(1+x\right)^{n-1}=C_n^1+2xC_n^2+...+n.x^{n-1}C_n^n\)
Thay \(x=1\)
\(\Rightarrow n.2^{n-1}=C_n^1+2C_n^2+...+nC_n^n\)
\(\Rightarrow n.2^{n-1}+1=C_n^0+C_n^1+2C_n^2+...+nC_n^n\)
\(\Rightarrow S=n.2^{n-1}+1\)
\(\Leftrightarrow\frac{1}{x}-\frac{2}{x\left(x+1\right)}=\frac{7}{6\left(x+4\right)}\)
\(\Leftrightarrow\frac{x-1}{x\left(x+1\right)}=\frac{7}{6\left(x+4\right)}\)
\(\Leftrightarrow6\left(x-1\right)\left(x+4\right)=7x\left(x+1\right)\)
\(\Leftrightarrow x^2-11x+24=0\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=8\end{matrix}\right.\)
\(\lim\limits_{x\rightarrow1}\dfrac{x^3-3x+2}{x^4-4x+3}=\lim\limits_{x\rightarrow1}\dfrac{\left(x+2\right)\left(x-1\right)^2}{\left(x^2+2x+3\right)\left(x-1\right)^2}=\lim\limits_{x\rightarrow1}\dfrac{x+2}{x^2+2x+3}=\dfrac{1}{2}\)
\(\lim\limits_{x\rightarrow2^-}\dfrac{x^3+x^2-4x-4}{x^2-4x+4}=\lim\limits_{x\rightarrow2^-}\dfrac{\left(x-2\right)\left(x^2+3x+2\right)}{\left(x-2\right)^2}=\lim\limits_{x\rightarrow2^-}\dfrac{x^2+3x+2}{x-2}=-\infty\)
\(\lim\limits_{x\rightarrow2}\dfrac{\left(x^2-x-2\right)^{20}}{\left(x^3-12x+16\right)^{10}}=\lim\limits_{x\rightarrow2}\dfrac{\left(x+1\right)^{20}\left(x-2\right)^{20}}{\left(x+4\right)^{10}\left(x-2\right)^{20}}=\lim\limits_{x\rightarrow2}\dfrac{\left(x+1\right)^{20}}{\left(x+4\right)^{10}}=\dfrac{3^{10}}{2^{10}}\)
\(\lim\limits_{x\rightarrow0^-}\dfrac{4x^2+5x}{x^2}=\lim\limits_{x\rightarrow0^-}\dfrac{4x+5}{x}=-\infty\)
\(\lim\limits_{x\rightarrow-1}\dfrac{\sqrt{x+2}-1}{\sqrt{x+5}-2}=\lim\limits_{x\rightarrow-1}\dfrac{\left(x+1\right)\left(\sqrt{x+5}+2\right)}{\left(x+1\right)\left(\sqrt{x+2}+1\right)}=\lim\limits_{x\rightarrow-1}\dfrac{\sqrt{x+5}+2}{\sqrt{x+2}+1}=2\)
\(S=\dfrac{1}{2018!\left(2019-2018\right)!}+\dfrac{1}{2016!\left(2019-2016\right)!}+...+\dfrac{1}{2!\left(2019-2\right)!}+\dfrac{1}{0!\left(2019-0!\right)}\)
\(\Rightarrow2019!.S=\dfrac{2019!}{2018!\left(2019-2018\right)!}+\dfrac{2019!}{2016!\left(2019-2016\right)!}+...+\dfrac{2019!}{2!\left(2019-2\right)!}+\dfrac{2019!}{0!\left(2019-0\right)!}\)
\(=C_{2019}^{2018}+C_{2019}^{2016}+...+C_{2019}^2+C_{2019}^0\)
\(=\dfrac{1}{2}\left(C_{2019}^0+C_{2019}^1+...+C_{2019}^{2018}+C_{2019}^{2019}\right)\)
\(=\dfrac{1}{2}.2^{2019}=2^{2018}\)
\(\Rightarrow S=\dfrac{2^{2018}}{2019!}\)
Chọn D