Cho dãy số (un) xác định bởi u(1)=0 và u(n+1)=u(n)+4n+3, với mọi n lớn hơn bằng 1. Biết lim căn bậc hai u(n)

Và

$\begin{array}{l}{u}_{4n}=2{\left(4n\right)}^2+4n-3\\ u_{4^{2}n}=2{\left(4^{2}n\right)}^2+4^{2}n-3\\ \mathrm{...}\\ u_{4^{2018}n}=2{\left(4^{2018}n\right)}^2+4^{2018}n-3\end{array}$

Do đó $\lim \frac{\sqrt{u_n}+\sqrt{u_{4n}}+\sqrt{u_{4^{2}n}}+\mathrm{...}+\sqrt{u_{4^{2018}n}}}{\sqrt{u_n}+\sqrt{u_{2n}}+\sqrt{u_{2^{2}n}}+\mathrm{...}+\sqrt{u_{2^{2018}n}}}$

$=\lim \frac{\sqrt{2+\frac{1}{n}-\frac{3}{n^2}}+\sqrt{{2.4}^2+\frac{4}{n}-\frac{3}{n^2}}+\mathrm{...}+\sqrt{2{\left(4^{2018}\right)}^2+\frac{4^{2018}}{n}-\frac{3}{n^2}}}{\sqrt{2+\frac{1}{n}-\frac{3}{n^2}}+\sqrt{{2.2}^2+\frac{2}{n}-\frac{3}{n^2}}+\mathrm{...}+\sqrt{2{\left(2^{2018}\right)}^2+\frac{2^{2018}}{n}-\frac{3}{n^2}}}$

$=\frac{\sqrt{2}\left(1+4+4^2+\mathrm{...}+4^{2018}\right)}{\sqrt{2}\left(1+2+2^2+\mathrm{...}+2^{2018}\right)}=\frac{1\frac{1-4^{2019}}{1-4}}{\frac{1-2^{2019}}{1-2}}=\frac{1}{3}\frac{4^{2019}-1}{2^{2019}-1}=\frac{2^{2019}+1}{3}$