[tex]\displaystyle 1+3+5+...+19= \\ \\ =1+2+3+4+5+...+19-(2+4+6+...+18)= \\ \\ = \frac{19(19+1)}{2} -2(1+2+3+...+9)= \frac{19 \times 20}{2} -2 \times \frac{9(9+1)}{2}= \\ \\ = \frac{380}{2} -2 \times \frac{9 \cdot 10}{2} =190- \not 2 \times \frac{90}{\not 2} =190-90=10(19-9)= \\ \\ =10 \times 10=10^2-p.p [/tex]