Question

verificati ca numarul  n=100+2X(1+2+3+....99)= este p.p

Answer 1

$\displaystyle 100+2 \cdot (1+2+3+...+99)=100+2 \cdot \frac{99(99+1)}{2} = \\ \\ =100+2 \cdot \frac{99 \cdot 100}{2} =100+ \not 2 \cdot \frac{9900}{\not 2} =100+9900= \\ \\ =100(1+99)=100 \cdot 100=100^2-p.p$