Question

Answer 1

\(u_{n+1}=u_n+4n+3\Rightarrow u_{n+1}-2\left(n+1\right)^2-\left(n+1\right)=u_n-2n^2-n\)

Đặt \(u_n-2n^2-n=v_n\Rightarrow\left\{{}\begin{matrix}v_1=0\\v_{n+1}=v_n=...=v_1=0\end{matrix}\right.\)

\(\Rightarrow u_n=2n^2+n\)

\(\Rightarrow lim\dfrac{\sqrt{u_n}+\sqrt{u_{4n}}+...+\sqrt{u_{4^{2018}n}}}{\sqrt{u_n}+\sqrt{u_{2n}}+...+\sqrt{u_{2^{2018}n}}}=\dfrac{\sqrt{2}\left(1+4+4^2+...+4^{2018}\right)}{\sqrt{2}\left(1+2+2^2+...+2^{2018}\right)}\)

\(=\dfrac{4^{2019}-1}{3}.\dfrac{1}{2^{2019}-1}=\dfrac{2^{2019}+1}{3}\)

\(\Rightarrow S=2+1-3=0\)