Question

Calculeaza urmatoarele sume:
a. 1+3+5+........+99=
b. 3+6+9+12+....+2001=
c. 4+8+12+....+2000=
d.5+10+15+....+2000=
e. 7+14+21+....+2002

Answer 1

$\displaystyle a).1+3+5+...+99= \\ \\ =1+2+3+4+5+...+99-(2+4+6+...+98)= \\ \\ = \frac{99(99+1)}{2} -2(1+2+3+...+49)= \frac{99 \times 100}{2} -2 \times \frac{49(49+1)}{2} = \\ \\ = \frac{9900}{2} -2 \times \frac{49 \times 50}{2} =4950- \not2 \times \frac{2450}{\not 2} =4950-2450=2500$

$\displaystyle b).3+6+9+12+...+2001=3(1+2+3+...+667)= \\ \\ =3 \times \frac{667(667+1)}{2} =3 \times \frac{667 \times 668}{2} = 3 \times \frac{445556}{2} = \\ \\ =3 \times222778=668334$

$\displaystyle c).4+8+12+...+2000=4(1+2+3+...+500)= \\ \\ =4 \times \frac{500(500+1)}{2} =4 \times \frac{500 \times 501}{2} =4 \times \frac{250500}{2} = \\ \\ =4 \times 125250=501000$

$\displaystyle d).5+10+15+...+2000=5(1+2+3+...+400)= \\ \\ =5 \times \frac{400(400+1)}{2} =5 \times \frac{400 \times 401}{2} =5 \times \frac{160400}{2} = \\ \\ =5 \times 80200=401000$

$\displaystyle e).7+14+21+...+2002=7(1+2+3+...+286)= \\ \\ =7 \times \frac{286(286+1)}{2} =7 \times \frac{286 \times 287}{2} =7 \times \frac{82082}{2} = \\ \\ =7 \times 41041=287287$