Question

(2+4+6+...+20) - (1+3+5+...+19)

Answer 1

2+4+6+...+20=110 1+3+5+...19=100
110-100=10
Sau 2+4+6+...+20-1-3-5-...-19 = 10
Grupam 2 cu -1, 4 cu -4,..., 20 cu -19 si vom obtine 1+1+....+1= 100 (1+1 de 100 de ori)

Answer 2

( 2 + 4 + 6 + .......+ 20 ) - ( 1 + 3 + 5 + ........+ 19 ) = 10

= 2 ( 1 + 2 + 3 + ........+ 10 ) - ( 10 × 20 / 2 ) =

= 2 × ( 10 × 11 / 2 ) - ( 10 × 10 ) =

= 2 × 55 - 100=

= 110 - 100 =

= 10

_______________

Cum am procedat :

( 2+ 4 + 6 + .....+ 20 ) =→ il dau factor comun pe 2

= 2 x ( 1 + 2 + 3 + ......+ 10 ) =

= 2 x [ 10 + ( 1 + 10 ) } / 2 → aplic formula sumei lui Gauss pentru suma primelor 10 numere consecutive

10 →termeni are suma

( 1 + 10 ) → primul adunat cu ultimul termen al sumei

→ apoi impart la 2

⇒ 2 x ( 10 × 11 ) / 2 = 110 → suma primei paranteze

a doua paranteza o rezolv aplicand suma lui Gauss

1 + 3 + 5 + ........+ 19 =

→ suma numerelor consecutive impare

→ stabilesc cati termeni are suma astfel: ( 19 + 1 ) : 2 = 20 : 2 = 10 termeni are suma

sau : ( 19 - 1 ) : 2 ( ratia) + 1 = 18 : 2 + 1 = 9 + 1 = 10

ratia este 2 : 3 - 1 ; 5 - 3;

= 10 × ( 1 + 19 ) / 2 = 10 × 20 / 2 = 100