Ta có:
\(1^2+3^2+...+99^2\)
\(=1^2+2^2+3^2+...+99^2-\left(2^2+4^2+...+98^2\right)\)
\(=1^2+2^2+3^2+...+99^2-4\left(1^2+2^2+...+49^2\right)\)
\(=\frac{99.\left(99+1\right)\left(2.99+1\right)}{6}-4.\frac{49.\left(49+1\right)\left(2.49+1\right)}{6}\)
\(=166650\)
\(1^2+3^{^2}+.......+99^2\) công thức tổng quát :\(1^2+3^2+5^2+....\) \(\left(2n-1\right)^2=\frac{n.\left(4n^2-1\right)}{3}\) suy ra A = \(1^2+3^2+5^2+.....+99^2\) TỪ 2n-1 =99 suy ra n =50 suy ra A =\(\frac{50.\left(4.50^2-1\right)}{3}\) A = 166650