Question

stabiliti paritatea sumelor:
S1: 2+4+6+...+198

Answer 1

2(1+2+...+99)=
=2* $\frac{99*100}{2}$ =
=99*100=
=9900.

Answer 2

2+4+6+...+198= 2( 1+ 2+ 3+...+99= 2 ( 99 X $\frac{100}{2}$)= 2 X 99 X 50= 9900
DECI REZULTATUL ESTE UN NUMAR PAR.