Question

Calculati sumele:

a) 4+8+12+......+1996
b)7+14+21+......+3157
c) 8+16+14+ ....+2504
d)11+22+33+.....+1311
e)19+38+57+...+1995

Answer 1

$\displaystyle a).4+8+12+...+1996=4(1+2+3+...+499)= \\ \\ =4 \times \frac{499(499+1)}{2} =4 \times \frac{499 \times 500}{2} =4 \times \frac{249500}{2} = \\ \\ =4 \times 124750=499000 \\ \\ b).7+14+21+...+3157=7(1+2+3+...+451)= \\ \\ =7 \times \frac{451(451+1)}{2} =7 \times \frac{451 \times 452}{2} =7 \times \frac{203852}{2} = \\ \\ =7 \times 101926=713482$

$\displaystyle c).8+16+24+...+2504=8(1+2+3+...+313)= \\ \\ =8 \times \frac{313(313+1)}{2} =8 \times \frac{313 \times 314}{2} =8 \times \frac{98282}{2} = \\ \\ =8 \times 49141=393128 \\ \\ d).11+22+33+...+1331=11(1+2+3+...+121)= \\ \\ =11 \times \frac{121(121+1)}{2} =11 \times \frac{121 \times 122}{2} =11 \times \frac{14762}{2} = \\ \\ =11 \times 7381=81191$

$\displaystyle e).19+38+57+...+1995=19(1+2+3+...+105)= \\ \\ =19 \times \frac{105(105+1)}{2} =19 \times \frac{105 \times 106}{2} =19 \times \frac{11130}{2} = \\ \\ =19 \times 5565=105735$