已知a=2013n+2010,b=2012n+2011,c=2013n+2012,求a^2+b^2+c^2-ab-bc-ac的值

已知a=2013n+2010,b=2012n+2011,c=2013n+2012
a-b=n-1，c-b=n+1，c-a=2

a^2+b^2+c^2-ab-bc-ac
=1/2*(2a^2+2b^2+2c^2-2ab-2bc-2ac)
=1/2*[(a-b)²+(c-b)²+(c-a)²]
=1/2*[(n-1)²+(n+1)²+2²]
=n²+3

∵a=2013n+2010,b=2013n+2011,c=2013n+2012
∴a-b=-1 a-c=-2 b-c=-1
a²+b²+c²-ab-bc-ac
=2(a²+b²+c²-ab-bc-ac)/2
=(2a²+2b²+2c²-2ab-2ac-2bc)/2
=(a²-2ab+b²+b²-2bc+c²+c²-2ac+a²)/2
=[(a-b)²+(b-c)²+(c-a)²]/2
=[(-1)²+(-1)²+(-2)²]/2
=(1+1+4)/2
=3

应该是b=2013n+2011
则a-b=-1
b-c=-1
c-a=2
所以原式=(2a²+2b²+2c²-2ab-2bc-2ac)/2
=[(a²-2ab+b²)+(b²-2bc+c²)+(c²-2ac+a²)]/2
=[(a-b)²+(b-c)²+(c-a)²]/2
=(1+1+4)/2
=3

a-b=n-1
b-c=1-n
a-c=-2
a^2+b^2+c^2-ab-bc-ac
=1/2*[(a-b)^2+(b-c)^2+(a-c)^2]
=1/2*[(n-1)^2+(1-n)^2+(-2)^2]
=1/2*(n^2-2n+1+n^2-2n+1+4)
=n^2-2n+3

你好
a^2+b^2+c^2-ab-bc-ac=（1/2）*2(a^2+b^2+c^2-ab-bc-ac)=（1/2）((a-b)^2+(b-c)^2+(a-c)^2)
因为a=2013n+2010,b=2012n+2011,c=2013n+2012
所以（a-b）^2=(2013n+2010-2012n-2011)^2=(n-1)^2
(b-c)^2=(n+1)^2
(a-c)^2=2^2
所以a^2+b^2+c^2-ab-bc-ac=（1/2）*（(2n-1)^2+(n+1)^2+4）=n^2+3

解：a-b=n-1 c-a=2 b-c=-1-n
所以原式=(2a²+2b²+2c²-2ab-2bc-2ac)/2
=[(a²-2ab+b²)+(b²-2bc+c²)+(c²-2ac+a²)]/2
=[(a-b)²+(b-c)²+(c-a)²]/2
={(n-1)^2+(-1-n)^2+4}/2
=n^2+3
正解，给个满意吧，谢谢