Question

Demonstreaza ca numarul a este divizibil cu 5,unde a=2+4+6+...+78

Answer 1

a=2+4+6+...+78

a=2(1+2+3+...+39)

(1+2+3+...+39)=39×40/2 pentru ca 1+2+3+....+n)=[n(n+1)]/2

a=2×(39×40/2)

a=39×40

a=1560 divizibil cu 5

Answer 2

a = 2 + 4 + 6 + .......+ 78

a₂ = 78 + 76 + 74 + ....... + 6 + 4 + 2

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2 x a = ( 2 + 78 ) + ( 4 + 76 ) + ( 6 + 74 ) + ......+ ( 74 + 6 ) + ( 76 + 4 ) + ( 78 + 2 )

2 x a = 80 + 80 + 80 + ..........+ 80 + 80

=> aflam cati termeni are suma

( 78 - 2 ) : 2 + 1 = 76 : 2 + 1 = 38 + 1 = 39 termeni are suma

2 x a = 80 x ( 1 + 1 + 1 + ......+ 1 ) ==> 80 este divizibil cu 5

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==> aplic formula sumei lui Gauss

a = 2 + 4 + 6 + 8 + .....+ 78

a = ( 78 - 2 ) : 2 + 1 = 76 : 2 + 1 = 38 + 1 = 39 termeni are suma

a = 39 x ( 2 + 78 ) / 2

a = 39 x 80 / 2

a =3120 / 2

a = 1560 ==> divizibil cu 5