Question

sa se afle 1+2+3-4+5+6+7-8+....2009+2010+2011-2012

Answer 1

Observam ca al patrulea termen din progresie se scade, deci vom avea:

1+2+3-4+5+6+7-8+...+2009+2010+2011-2012=
(1+2+3+4+5+...+2009+2010+2011+2012)-4-8-12-...-2012=2012(2012+1)/2 -(4+8+12+...+2012)=
1006·2013-4(1+2+3+...+503)=1006·2013-4·503(503+1)/2=1006·2013-1006·504=1006(2013-504)=... calculezi...

S-a utilizat Suma Gauss:
1+2+3+..+n=n(n+1)/2

Answer 2

   
 [tex]\displaystyle \\ \text{Avem sirul:} \\ 1+2+3-4+5+6+7-8+\cdots+2009+2010+2011-2012 \\ \text{Voi evidentia inca 8 termeni:} \\ 1+2+3-4+5+6+7-8+9+10+11-12+ \\ +13+14+15-16+ \cdots +2009+2010+2011-2012 \\ \text{Observam ca dupa fiecare 3 termeni pozitivi urmeaza unu negativ.} \\ \text{Vom grupa termenii sirului in grupe de 4 termeni } \\ \text{si adunam termenii din fiecare grupa.}[/tex]
 
 
[tex]\displaystyle \\ (1+2+3-4)+(5+6+7-8)+(9+10+11-12)+ \\ +(13+14+15-16)+ \cdots +(2009+2010+2011-2012) =\\ \\ = 2 + 10+18+26+ \cdots +4018 \\ \\ \text{Am obtinut un sir Gauss cu ratia = 8, pe care il putem calcula usor.} \\ \\ \text{Aflam numarul de termeni:} \\ \\ n = \frac{4018-2}{8}+1 = \frac{4016}{8}+1 = 502+1 = 503 ~\text{termeni} \\ \\ 2 + 10+18+26+ \cdots +4018 = \frac{503(4018+2)}{2}= \\ \\ \frac{503\times 4020}{2}= 503 \times 2010 = \boxed{1011030}[/tex]