Question

Să se calculeze sumele :
a.) 1+3+5+...+2011
b.) 2005+2+4+6+...+4008
c.) 1996+2+4+6+...+3990
d.) 22+24+26+...+498 .
Vă rog , să-mi spune-ți cum ați făcut . ( mai explicit )

Answer 1

1)n=(2011-1):r+1

n-nr de elemente din sir
r-ratia

r=3-1=2

n=2010:2+1

n=1005+1
n=1006

S=(1+2011)*1006/2= 2012*503=  503*4 *503=  503² *4 = 1 012 036

b)2005 +2(1+2+3+...+2004)=  2005+2(1+2004)*2004/2= 2005+(1+2004)*2004=

=2005+2005*2004=  2005(1+2004)= 2005*2005=  2005²

c)1996+2(1+2+3+....+1995)=1996 +2(1+1995)*1995/2=  1996 +(1995+1)*1995=

=1996+1996*1995=  1996(1+1995)= 1996*1996=  1996²

d)2(11+12+13+...+249)= 2*(11+249)*239/2= 260*239=62140