Question

determinati suma numerelor care impartite la 4 dau restul 3

Answer 1

numerele de forma 4n+3
n=0 => 4n+3=3
n=1 => 4n+3=7
n=2 => 4n+3=11
....
n=x => 4x+3
suma lor este:
3    +    7+     11+.....+4x+3=(0+3)+(1*4+3)+(2*4+3)+......+(x*4+3)=
=0+4+2*4+3*4+.....+x*4+3*(x+1)=
=4(1+2+3+....+x)+3*(x+1)
=4*x*(x+1)/2+3*(x+1)=2*x*x+2x+3x+3=
$=2 x^{2} +5x+3$