Lời giải:Ta có:
\(\begin{array}{l}\,\,\,\,\lim \left( {\frac{{{{\left( {\sqrt 5 } \right)}^n} - {2^{n + 1}} + 1}}{{{{5.2}^n} + {{\left( {\sqrt 5 } \right)}^{n + 1}} - 3}} + \frac{{2{n^2} + 3}}{{{n^2} - 1}}} \right)\\ = \lim \frac{{{{\left( {\sqrt 5 } \right)}^n} - {2^{n + 1}} + 1}}{{{{5.2}^n} + {{\left( {\sqrt 5 } \right)}^{n + 1}} - 3}}\\ + \lim \frac{{2{n^2} + 3}}{{{n^2} - 1}}\\ = \lim \frac{{1 - {{\left( {\frac{2}{{\sqrt 5 }}} \right)}^n}.2 + {{\left( {\frac{1}{{\sqrt 5 }}} \right)}^n}}}{{5.{{\left( {\frac{2}{{\sqrt 5 }}} \right)}^n} + \sqrt 5 - {{\left( {\frac{3}{{\sqrt 5 }}} \right)}^n}}}\\ + \lim \frac{{2 + \frac{3}{{{n^2}}}}}{{1 - \frac{1}{{{n^2}}}}}\\ = \frac{{1 - 2.0 + 0}}{{5.0 + \sqrt 5 - 0}} + \frac{2}{1}\\ = \frac{{\sqrt 5 }}{5} + 2\end{array}\)
\( \Rightarrow a = 1,\,\,b = 5,\,\,c = 2\).
Vậy \(S = {a^2} + {b^2} + {c^2} = {1^2} + {5^2} + {2^2} = 30.\)
