\(u_{n+1}=u_n+4n+3\)
\(\Leftrightarrow u_{n+1}-2\left(n+1\right)^2-\left(n+1\right)=u_n-2n^2-n\)
Đặt \(v_n=u_n-2n^2-n\Rightarrow\left\{{}\begin{matrix}v_1=u_1-2-1=-3\\v_{n+1}=v_n=...=v_1=-3\end{matrix}\right.\)
\(\Rightarrow u_n-2n^2-n=-3\Rightarrow u_n=2n^2-n-3\)
\(\Rightarrow\sqrt{u_{4^kn}}=\sqrt{2\left(4^kn\right)^2-4^kn-3}=\sqrt{2}.4^kn.\sqrt{1-\dfrac{1}{2.4^kn}-\dfrac{3}{2.\left(4^kn\right)^2}}\)
Tương tự: \(\sqrt{u_{2^mn}}=\sqrt{2}.2^mn.\sqrt{1-\dfrac{2}{2.4^mn}-\dfrac{3}{2.\left(4^mn\right)^2}}\)
Do đó:
\(\lim...=\lim\dfrac{\sqrt{2}\left(1+4+4^2+...+4^{2018}\right)}{\sqrt{2}\left(1+2+2^2+...+2^{2018}\right)}=\dfrac{4^{2019}-1}{3}.\dfrac{1}{2^{2019}-1}\)
\(=\dfrac{\left(2^{2019}\right)^2-1}{3\left(2^{2019}-1\right)}=\dfrac{\left(2^{1029}-1\right)\left(2^{2019}+1\right)}{3}=\dfrac{2^{2019}+1}{3}\)
\(\Rightarrow a=2;b=1;c=3\)
