Lời giải của Tự Học 365

Xét $g\left( n \right) = \dfrac{{f\left( {2n - 1} \right)}}{{f\left( {2n} \right)}} \Rightarrow g\left( n \right) = \dfrac{{{{\left( {4{n^2} - 2n + 1} \right)}^2} + 1}}{{{{\left( {4{n^2} + 2n + 1} \right)}^2} + 1}}$.

$g\left( n \right) = \dfrac{{{{\left( {4{n^2} + 1} \right)}^2} - 4n\left( {4{n^2} + 1} \right) + \left( {4{n^2} + 1} \right)}}{{{{\left( {4{n^2} + 1} \right)}^2} + 4n\left( {4{n^2} + 1} \right) + \left( {4{n^2} + 1} \right)}} = \dfrac{{4{n^2} + 1 - 4n + 1}}{{4{n^2} + 1 + 4n + 1}} = \dfrac{{{{\left( {2n - 1} \right)}^2} + 1}}{{{{\left( {2n + 1} \right)}^2} + 1}}$

$\Rightarrow {u_n} = \dfrac{2}{{10}}.\dfrac{{10}}{{26}}.\dfrac{{26}}{{50}}....\dfrac{{{{\left( {2n - 3} \right)}^2} + 1}}{{{{\left( {2n - 1} \right)}^2} + 1}}.\dfrac{{{{\left( {2n - 1} \right)}^2} + 1}}{{{{\left( {2n + 1} \right)}^2} + 1}} = \dfrac{2}{{{{\left( {2n + 1} \right)}^2} + 1}}$

$\Rightarrow \lim n\sqrt {{u_n}}  = \lim \sqrt {\dfrac{{2{n^2}}}{{4{n^2} + 4n + 2}}}  = \dfrac{1}{{\sqrt 2 }}.$

Đáp án cần chọn là: d