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42 votes
42 votes
1+(-2)+3+(-4)+...+2011+(-2012)+2013

User DritanX
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1 Answer

25 votes
25 votes

Answer:

1007

Explanation:

Divide the expression by two parts: the first one with all positive integer (the integer which can't be divided by two), 1+3+5+...2013 and the rest -2-4-6-...-2012

If you notice both parts are arithmetical sequences

Explore the first one 1+3+5...+2013

a1=1 d=3-1=2 an=2013

a1+d(n-1)=an

1+2(n-1)=2013

2(n-1)=2012

n-1=1006

n=1007

Sn=( 1/2)*(a1+an)*n= 0.5*2014*1007=1007*1007=1014049

-2+(-4)-...-2012

a1=-2

d=a2-a1= -4-(-2)=-2

a1+d(n-1)=an

-2+(-2)(n-1)=-2012

n=1006

Sn= (a1+an)/2*n=(-2-2012)/2*1006=-1007*1006

-1007*1006+1007*1007= 1007- the answer

The second mean

1-2=-1

3-4=-1

5-6=-1

There are 2012/2=1006 couples and every of them has sum -1

-1*1006=-1006

-1006+2013=1007

User Zongshiwujie
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