Question

Repede Pls va las poza,trebuie prin metoda lui gauss ex 2

Answer 1

a.) S= n(n+1)/2= 201(201+1)/2= (201×202)/2= 40602/2= 20301

b.) S= n(n+1)/2= 306(306+1)/2= (306×307)/2= 93942/2= 46971

c.) S= n(n+1)/2= 1999(1999+1)/2= (1999×2000)/2= 3998000/2= 1999000

"n" => ultimul termen al sumei

Answer 2

$a)1 + 2 + 3 + ... + 201$

$= \frac{201(201 + 1)}{2} = \frac{201 \times 202}{2} = 201 \times 101 = 20301$

$b)1 + 2 + 3 + ... + 306$

$= \frac{306(306 + 1)}{2} = \frac{306 \times 307}{2} = 153 \times 307 = 46971$

$c)1 + 2 + 3 + ... + 1999$

$= \frac{1999(1999 + 1)}{2} = \frac{1999 \times 2000}{2} = 1999 \times 1000 = 1999000$

Formulă :

$1 + 2 + 3 + ... + n = \frac{n(n + 1)}{2}$