Răspuns :
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
[tex]a)1 + 2 + 3 + ... + 201[/tex]
[tex] = \frac{201(201 + 1)}{2} = \frac{201 \times 202}{2} = 201 \times 101 = 20301[/tex]
[tex]b)1 + 2 + 3 + ... + 306[/tex]
[tex] = \frac{306(306 + 1)}{2} = \frac{306 \times 307}{2} = 153 \times 307 = 46971[/tex]
[tex]c)1 + 2 + 3 + ... + 1999[/tex]
[tex] = \frac{1999(1999 + 1)}{2} = \frac{1999 \times 2000}{2} = 1999 \times 1000 = 1999000[/tex]
Formulă :
[tex]1 + 2 + 3 + ... + n = \frac{n(n + 1)}{2} [/tex]