Răspuns :

Iakabcristina2id

Răspuns:

1. Este corect.

2. 2(1+2+3+...+50) = 2 × (50×51)/2 = 2550

3. 1+2(n-1) =101 =>n=51

1+3+5+...+101 = 51(1+101)/2 = 51×51 = 2601

4. 39+40+41+...110 = (110×111)/2 - (38×39)/2 = 6105 - 741 = 5364

5. 5+10+15+20+...+2020 = 5(1+2+3+...+404) = 5(404×405)/2 = 5×202×405 = 409050

Dianageorgiana794id

Răspuns:

1) formula lui Gauss

1 + 2 + 3 + 4 + … + n = n·( n + 1 ) : 2

2) 2(1+2+3+...+50) = 2·[50·(50+1)/2 =50·51=2550

dai factor comun 2

3) 1+2(n-1) =101 =>n=51

Formula lui Gauss pentru sume de numere impare

1 + 3 + 5 + 7 + … + ( 2n-1 ) = n·n

1+3+5+...+101 = 51·(1+101)/2=51·51 =2601

4)  39+40+..+100=(1+2+3+...+110)-(1+2+3+4+..+38)

=110·111:2-38·39:2 =6105-741=5364

5) 5+10+15+20+...+2020  

dam factor comun 5

5(1+2+3+...+404) =5·(404·405)/2 =5·404·405:2=409050

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