Ta có:

\(\begin{array}{l}C_n^2C_n^{n - 2} + C_n^8C_n^{n - 8} = 2C_n^2C_n^{n - 8} \Leftrightarrow {\left( {C_n^2} \right)^2} - 2C_n^2C_n^8 + {\left( {C_n^8} \right)^2} = 0\\ \Leftrightarrow {\left( {C_n^2 - C_n^8} \right)^2} = 0 \Leftrightarrow C_n^2 - C_n^8 = 0 \Leftrightarrow \dfrac{{n!}}{{2!.\left( {n - 2} \right)!}} - \dfrac{{n!}}{{8!.\left( {n - 8} \right)!}} = 0\\ \Leftrightarrow \dfrac{1}{{\left( {n - 2} \right)\left( {n - 3} \right)\left( {n - 4} \right)\left( {n - 5} \right)\left( {n - 6} \right)\left( {n - 7} \right)}} - \dfrac{1}{{8.7.6.5.4.3}} = 0\\ \Leftrightarrow \left( {n - 2} \right)\left( {n - 3} \right)\left( {n - 4} \right)\left( {n - 5} \right)\left( {n - 6} \right)\left( {n - 7} \right) = 8.7.6.5.4.3 \Leftrightarrow n = 10\end{array}\)

Xét hàm số: \(f\left( x \right) = {(x + 1)^{10}} = \sum\limits_{i = 0}^{10} {C_{10}^i{x^i}} \) có:

\(\begin{array}{l}f'\left( x \right) = 10{(x + 1)^9} = \sum\limits_{i = 0}^{10} {C_{10}^ii{x^{i - 1}}} \\ \Rightarrow x.f'\left( x \right) = 10x{(x + 1)^9} = \sum\limits_{i = 0}^{10} {C_{10}^ii{x^i}} \\ \Rightarrow {\left( {x.f'\left( x \right)} \right)^\prime } = {\left( {10x{{(x + 1)}^9}} \right)^\prime } = \sum\limits_{i = 0}^{10} {C_{10}^i{i^2}{x^{i - 1}}} \\ \Rightarrow 10x.9{(x + 1)^8} + 10{(x + 1)^9} = \sum\limits_{i = 0}^{10} {C_{10}^i{i^2}{x^{i - 1}}} \\ \Rightarrow 90{x^2}{(x + 1)^8} + 10x{(x + 1)^9} = \sum\limits_{i = 0}^{10} {C_{10}^i{i^2}{x^i}} \\ \Rightarrow \sum\limits_{i = 0}^{10} {C_{10}^i{i^2}{x^i}}  = 90{x^2}{(x + 1)^8} + 10x{(x + 1)^9}\end{array}\)

\(\begin{array}{l} \Rightarrow T = {1^2}C_n^1 + {2^2}C_n^2 + ... + {n^2}C_n^2\\\,\,\,\,\,\,\,\,\,\,\, = {1^2}C_{10}^1 + {2^2}C_{10}^2 + ... + {10^2}C_{10}^{10}\\\,\,\,\,\,\,\,\,\,\,\, = {90.1.2^8} + {10.1.2^9} = {55.2^9}\end{array}\) (ứng với \(x = 1\)).

Chọn: A