Ta có công thức công thức tính tổng bình phương các số lẻ từ 1 đến 2m-1 là:
\(1^2+3^2+\cdots+\left(2m-1\right)^2\)
\(=1^2+2^2+\cdots+\left(2m\right)^2-\left\lbrack2^2+4^2+\cdots+\left(2m\right)^2\right\rbrack\)
\(=\frac{2m\left(2m+1\right)\left(2\cdot2m+1\right)}{6}-2^2\left(1^2+2^2+\cdots+m^2\right)\)
\(=\frac{m\left(2m+1\right)\left(4m+1\right)}{3}-4\cdot\frac{m\left(m+1\right)\left(2m+1\right)}{6}\)
\(=\frac{m\left(2m+1\right)\left(4m+1\right)}{3}-\frac{2m\left(m+1\right)\left(2m+1\right)}{3}\)
\(=\frac{m\left(2m+1\right)\left(4m+1\right)-2m\left(m+1\right)\left(2m+1\right)}{3}=\frac{\left(2m+1\right)\left(4m^2+m-2m^2-2m\right)}{3}\)
\(=\frac{\left(2m+1\right)\left(2m^2-m\right)}{3}=\frac{m\left(2m-1\right)\left(2m+1\right)}{3}\)
\(11^2+13^2+\cdots+2009^2\)
\(=\left(1^2+3^2+\cdots+2009^2\right)-\left(1^2+3^2+\cdots+9^2\right)\)
\(=\left\lbrack1^2+3^2+\cdots+\left(2\cdot1010-1\right)^2\right\rbrack-\left\lbrack1^2+3^2+\cdots+\left(2\cdot5-1\right)^2\right\rbrack\)
\(=\frac{1010\left(2\cdot1010-1\right)\left(2\cdot1010+1\right)}{3}-\frac{5\left(2\cdot5-1\right)\left(2\cdot5+1\right)}{3}\)
\(=1373734330-165=1373734165\)
