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