Question

a)S1=1+3+5...?2019=
b) S2=4+8+12...+2020=
c)S3=4+8+12+...+2020

VĂ ROG DAU COROANĂ​

Answer 1

Hei! :)

$S1=1+3+5...+2019\\=> ( 2019 - 1 ) : 2 + 1 = 1009 + 1=1010 (termenii\ sumei\ numerelor\ impare)\\=> S1=1010 *( 1 + 2019) : 2=1010 * 1010 = 1 020 100$

Observ ca S2 si S3 sunt aceeasi suma.

$S2=S3=4 + 8 + 12 + .....+ 2020= 4 * ( 1 + 2 + 3 + ....... + 505 )\\=4 * 505 * ( 1+505) : 2 =\\=2 * 505 * 506 =\\=1010 * 506 =\\=511 060$