Question

Aflati x din egalitatile:
a. 1+3+5+7+....+2009+2011+x=2+4+6+8+...+2010+2012

Answer 1

Notam suma de la 1 la 2011 cu y
y= (2011+1)*1006:2= 2012*1006:2= 1006²

Notam suma de la 2 la 2012 cu Z
Z= (2012+2)*1006:2= 2014*1006:2= 1007*1006
In ambele cazuri 1006 reprezinta numarul de termeni ai sumei, formula este Nr termeni+( ultimul-primul):diferenta+1
Formula pt cele 2 sume este : S=(primul+ultimul)*nr termeni:2

1006*1006+x= 1007*1006⇒ x=1006

Answer 2

1+3+5+7+....+2009+2011=(2011+1)x(numarul de termeni):2
numarul de termeni=(2011-1):2+1=2010:2+1=1005+1=1006
(2011+1)x10006:2=2012x1006:2=2024072:2=1012036
2+4+6+8+......+2010+2012=(2012+2)x(numarul de termeni):2
numarul de termeni=(2012-2):2+1=2010:2+1=1005+1=1006
(2012+2)x1006:2=2014x1006:2=2026084:2=1013042
1012036+x=1013042
x=1013042-1012036
x=1006