Question

在数列｛an｝中,a1=t,a2=t＾2(t大于0),且a(n+1)=(t+1)an-ta(n-1),(n大于等于2)

Answer 1

a(n+1)=(t+1)an-ta(n-1),

a(n+1)- an=t(an- a(n-1)),

{ an- a(n-1)}是等比数列,

an- a(n-1)=( t＾2-t)t^(n-2),

an=a1+(a2-a1)+……+( an- a(n-1))

=t+( t＾2-t)+ ( t＾2-t)t+……+( t＾2-t)t^(n-2)

=t^n.