已知a=2/1(二分之一)-2n+1,b=2/1(二分之一)-2n+3,c=2/1(二分之一)-2n+2,求a2(a的平

已知a=2/1(二分之一)-2n+1,b=2/1(二分之一)-2n+3,c=2/1(二分之一)-2n+2,求a2(a的平方)+b2(b的平方)+c2(c的平方)-ab-ac-bc的植最简便.
a^2+b^2+c^2-ab-ac-bc
=1/2[2a^2+2b^2+2c^2-2ab-2ac-2bc]
=1/2[(a-c)^2+(a-b)^2+(b-c)^2]
=1/2[(1/2-2n+1-1/2+2n-2)^2+(1/2-2n+1-1/2+2n-3)^2+(1/2-2n+3-1/2+2n-2)^2]
=1/2[1+4+1]
=3
a+1)(2a-3)-(a-1)(2a+3)
=[a+1][2a-3-2a-3]
=-6[a+1]

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