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Lời giải:
\(\frac{(x^2+x+1)^{2018}+(x+2)^{2018}-2.3^{2018}}{(x-1)(x+2017)}=\frac{(x^2+x+1)^{2018}-3^{2018}+(x+2)^{2018}-3^{2018}}{(x-1)(x+2017)}\)
\(=\frac{(x^2+x-2)[(x^2+x+1)^{2017}+...+3^{2017}]+(x-1)[(x+2)^{2017}+...+3^{2017}]}{(x-1)(x+2017)}\)
\(=\frac{(x+2)[(x^2+x+1)^{2017}+...+3^{2017}]+(x+2)^{2017}+...+3^{2017}}{x+2017}\)
Do đó:
\(\lim_{x\to 1}\frac{(x^2+x+1)^{2018}+(x+2)^{2018}-2.3^{2018}}{(x-1)(x+2017)}=\lim_{x\to 1}\frac{(x+2)[(x^2+x+1)^{2017}+...+3^{2017}]+(x+2)^{2017}+...+3^{2017}}{x+2017}\)
\(=\frac{3\underbrace{(3^{2017}+3^{2017}+...+3^{2017})}_{2018}+\underbrace{3^{2017}+...+3^{2017}}_{2018}}{1+2017}\)
\(=\frac{3.2018.3^{2017}+2018.3^{2017}}{2018}=3^{2018}+3^{2017}=3^{2017}.4\)
Xét khai triển:
\(\left(1+x\right)^{2017}=C_{2017}^0+xC_{2017}^1+x^2C_{2017}^2+...+x^{2017}C_{2017}^{2017}\)
Lấy tích phân 2 vế:
\(\int\limits^1_0\left(1+x\right)^{2017}=\int\limits^1_0\left(C_{2017}^0+xC_{2017}^1+...+x^{2017}C_{2017}^{2017}\right)\)
\(\Leftrightarrow\dfrac{2^{2018}-1}{2018}=C_{2017}^0+\dfrac{1}{2}C_{2017}^1+...+\dfrac{1}{2018}C_{2017}^{2017}\)
Vậy \(S=\dfrac{2^{2018}-1}{2018}\)
\(u_{n+1}^2=\dfrac{u_n^2}{1+u_n^2}\Rightarrow\dfrac{1}{u_{n+1}^2}=\dfrac{1}{u_n^2}+1\)
Đặt \(\dfrac{1}{u_n^2}=v_n\Rightarrow\left\{{}\begin{matrix}v_1=\dfrac{1}{2018^2}\\v_{n+1}=v_n+1\end{matrix}\right.\)
\(v_n\) là cấp số cộng với công sai d=1 \(\Rightarrow v_n=\dfrac{1}{2018^2}+n-1\)
\(\Rightarrow u_n^2=\dfrac{1}{v_n}=\dfrac{1}{n+\dfrac{1}{2018^2}-1}\)
\(u_n^2< \dfrac{1}{2018^2}\Rightarrow\dfrac{1}{n+\dfrac{1}{2018^2}-1}< \dfrac{1}{2018^2}\Rightarrow n...\)
