Question

efectuati a) 3+6+9+......+33 b)10+20+30+.......+990

Answer 1

$\displaystyle a)3+6+9+...+33=3(1+2+3+...+11)=3 \cdot \frac{11(11+1)}{2} = \\ \\ =3 \cdot \frac{11 \cdot 12}{2} =3 \cdot 11 \cdot 6=33 \cdot 6=198 \\ \\ b)10+20+30+...+990=10 (1+2+3+...+99)=10 \cdot \frac{99(99+1)}{2} =\\ \\ =5 \cdot 99 \cdot 100=495 \cdot 100=49500$