Question

2+4+6+...+158=?
3+6+9+...+162=?
4+8+12+...+288=?
7+14+21+...+147=?

Answer 1

$\displaystyle a).2+4+6+...+158=2(1+2+3+...+79)=2 \times \frac{79(79+1)}{2} = \\ \\ =2 \times \frac{79 \times 89}{2} =\not 2 \times \frac{6320}{\not2} =6320 \\ \\ b).3+6+9+...+162=3(1+2+3+...+54)=3 \times \frac{54(54+1)}{2} = \\ \\ =3 \times \frac{54 \times 55}{2} =3 \times \frac{2970}{2} =3 \times 1485=4455$

$\displaystyle c).4+8+12+...+288=4(1+2+3+...+72)=4 \times \frac{72(72+1)}{2} = \\ \\ =4 \times \frac{72 \times 73}{2} =4 \times \frac{5256}{2} =4 \times 2628=10512 \\ \\ d).7+14+21+...+147=7(1+2+3+...+21)=7 \times \frac{21(21+1)}{2} = \\ \\ =7 \times \frac{21 \times 22}{2} =7 \times \frac{462}{2} =7 \times 231=1617$