Sn=(1/4)(An)^2+(1/2)An
S(n-1)=(1/4)[A(n-1)]^2+(1/2)A(n-1)
两式相减:
an=Sn-S(n-1)
=(1/4)(An)^2+(1/2)An-{(1/4)[A(n-1)]^2+(1/2)A(n-1)}
an=(1/4)(An)^2+(1/2)An-(1/4)[A(n-1)]^2-(1/2)A(n-1)
(An)^2-2An-[A(n-1)]^2-2A(n-1)=0
(An)^2-[A(n-1)]^2=2[An+A(n-1)]
因An+A(n-1)≠0
An-A(n-1)=2
An=A1+2(n-1)
又S1=(1/4)(A1)^2+(1/2)A1
A1=(1/4)(A1)^2+(1/2)A1
(A1)^2-2A1=0
A1=2
An=2n