[tex]1+3+5+...+n = (\frac{n+1}{2})^{2} \\ 1+2+3+...+n = \frac{n(n+1)}{2} \\ \\ (1+3+5+7+...+99)\cdot(2+4+6+8+...+.98) = \\ \\ = (\frac{99+1}{2}) ^{2} \cdot 2\cdot(1+2+3+4+...+98) = \frac{10000}{4} \cdot2\cdot \frac{98\cdot99}{2} = \\ \\ =2500\cdot98\cdot99 = 24255000 [/tex]
