(×-1)(×-2)分之3×-4=(×-1)分之A+(×-2)分之B,求A,B

(×-1)(×-2)分之3×-4=(×-1)分之A+(×-2)分之B,求A,B
(×-1)(×-2)分之3×-4=(×-1)分之A+(×-2)分之B,求A,B

(×-1)(×-2)分之3×-4=(×-1)分之A+(×-2)分之B,求A,B
右边通分=[a(x-2)+b(x-1)]/(x-1)(x-2)
=[(a+b)x-(2a+b)]/(x-1)(x-2)
所以3x-4=(a+b)x-(2a+b)
a+b=3
2a+b=4
所以a=1,b=2

累加是an-a1=1+2+……+（n-1)=n(n-1)/2
a1=1
所以an=n(n-1)/2+1
=(3+i)(1+i)/[(1-i)(1+i)]
=(3+3i+i-1)/(1+1)
=(2+4i)/2
=1+2i