Question

Care este suma primelor 63 de numere naturale consecutive impare?

Answer 1

1 + 3 + 5 +... + n

termen 1 -> 1 = 2 × 1 - 1
termen 2-> 3 = 2 × 2 - 1
termen 3 -> 5 = 2 × 3 - 1
......
termen 63 -> 2 ×63 - 1 = 125

1 + 3 + 5 +...+ 125 =

= (125 + 1) × 63 : 2 =

= 126 × 63 : 2 =

= 63 × 63 =

= 3 969

Answer 2

63×2=126
deci ultimul Nr impar este 125
S=1+3+5+.....+125
S=125+123+......+1
adunând sumele
2S=126+126+.....+126
2S=126×63
S=63×63
S=3969