Đáp án:

\({1^2} + {3^2} + {5^2} + ...... + {\left( {2n - 1} \right)^2} = \dfrac{{n\left( {2n - 1} \right)\left( {2n + 1} \right)}}{3}.\)

Giải thích các bước giải:

\(\begin{array}{l}
\text{Đặt: }S\left( {2n - 1} \right) = {1^2} + {3^2} + {5^2} + ...... + {\left( {2n - 1} \right)^2}\\
S\left( {2n} \right) = {2^2} + {4^2} + {6^2} + ..... + {\left( {2n} \right)^2} = \dfrac{{4n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}\\
 \Rightarrow S\left( {2n - 1} \right) + S\left( {2n} \right) = {1^2} + {2^2} + {3^2} + {4^2} + ...... + {\left( {2n - 1} \right)^2} + {\left( {2n} \right)^2}\,\left( * \right)\\
\text{Ta có: }{1^2} + {2^2} + ...... + {n^2} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}\\
 \Rightarrow \left( * \right) \Leftrightarrow S\left( {2n - 1} \right) + S\left( {2n} \right) = \dfrac{{\left( {2n} \right)\left( {2n + 1} \right)\left( {4n + 1} \right)}}{6}.\\
 \Leftrightarrow S\left( {2n - 1} \right) = \dfrac{{\left( {2n} \right)\left( {2n + 1} \right)\left( {4n + 1} \right)}}{6} - \dfrac{{4n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}\\
 \Leftrightarrow S\left( {2n - 1} \right) = \dfrac{{2n\left( {2n + 1} \right)}}{6}\left( {4n + 1 - 2n - 2} \right) = \dfrac{{n\left( {2n - 1} \right)\left( {2n + 1} \right)}}{3}.\\
 \Rightarrow {1^2} + {3^2} + {5^2} + ... + {\left( {2n - 1} \right)^2} = \dfrac{{n\left( {2n - 1} \right)\left( {2n + 1} \right)}}{3}.
\end{array}\)