Tính tổng (1+1/3)^2 +(9+1/9)^2 +...+(3^n+1/3^n)^2

$\begin{array}{l}
S = {\left( {3 + \dfrac{1}{3}} \right)^2} + {\left( {9 + \dfrac{1}{9}} \right)^2} + ... + {\left( {{3^n} + \dfrac{1}{{{3^n}}}} \right)^2}\\
 = \underbrace {\left( {{3^2} + {3^4} + {3^6} + ... + {3^{2n}}} \right)}_A + \underbrace {\left( {\dfrac{1}{{{3^2}}} + \dfrac{1}{{{3^4}}} + \dfrac{1}{{{3^6}}} + ... + \dfrac{1}{{{3^{2n}}}}} \right)}_B\underbrace { + \underbrace {2 + 2 + ... + 2 + 2}_n}_C\\
 = A + B + C\\
A = {3^2} + {3^4} + {3^6} + ... + {3^{2n}}\\
9A = {3^4} + {3^6} + ... + {3^{2n + 2}}\\
 \Rightarrow 8A = {3^{2n + 2}} - {3^2} \Rightarrow A = \dfrac{{{3^{2n + 2}} - {3^2}}}{8}\\
B = \dfrac{1}{{{3^2}}} + \dfrac{1}{{{3^4}}} + ... + \dfrac{1}{{{3^{2n - 2}}}} + \dfrac{1}{{{3^{2n}}}}\\
9B = 1 + \dfrac{1}{{{3^2}}} + ... + \dfrac{1}{{{3^{2n - 2}}}}\\
 \Rightarrow 9B - B = 1 - \dfrac{1}{{{3^{2n}}}}\\
 \Rightarrow B = \dfrac{{1 - \dfrac{1}{{{3^{2n}}}}}}{8}\\
C = 2n\\
 \Rightarrow S = \dfrac{{{3^{2n + 2}} - {3^2}}}{8} + \dfrac{{1 - \dfrac{1}{{{3^{2n}}}}}}{8} + 2n
\end{array}$