Răspunsuri

  • Utilizator Brainly
2014-03-04T07:14:25+02:00
A) 1+2+3+...+200=200·201/2=20100;
se aplica formula: 1+2+3+...+n= \frac{n(n+1)}{2}

b) 1^{2} + 2^{2}+ 3^{2} +...+ 2014^{2}= \frac{2014*2015*4029}{6} ;
se aplica formula:  1^{2}+ 2^{2}+ 3^{2}+...+ n^{2}= \frac{n(n+1)(2n+1)}{6}

Ambele sunt formule de nivel gimnaziu.
2014-03-04T13:21:12+02:00
Folosesti 
1+2+...+n= \frac{n(n+1)}{2} 

1^{2} +  2^{2} +...+ n^{2} =  \frac{n(n+1)(2n+1)}{6}
poate la alte exercitii vei avea nevoie si de
 1^{3} + 2^{3} +...+ n^{3} = (\frac{n(n+1)}{2})^{2}