\(A=1^2+3^2+...+99^2\)
\(=\left(1^2+2^2+3^2+...+99^2\right)-\left(2^2+4^2+...+98^2\right)\)
\(=\dfrac{99\left(99+1\right)\left(2\cdot99+1\right)}{6}-2^2\left(1^2+2^2+...+49^2\right)\)
\(=\dfrac{99\left(99+1\right)\left(2\cdot99+1\right)}{6}-2^2\cdot\dfrac{49\left(49+1\right)\left(2\cdot49+1\right)}{2}\)
\(=328350-2^2\cdot40425=166650\)