已知数列{an},{bn}与函数f(x),g(x),x∈R满足条件:
2个回答

由条件知：tbn+1=2b(n+1),且t≠2.可得

b(n+1)+1/(t-2)=(t/2)[bn+1/(t-2)].

由f(b)≠g(b),t≠2,t≠0,可知b+1/(t-2)≠0,t/2≠0,

所以{bn+1/(t-2)}是首项为b+1/(t-2),公比为t/2的等比数列.

bn+1/(t-2)=[b+1/(t-2)](t/2)^(n-1),即bn=[b+1/(t-2)](t/2)^(n-1)-1/(t-2)

由an=2b(n+1)可知,若liman(n→∞）存在,则limbn(n→∞）存在.

于是可得0＜|t/2|＜1,故-2＜t＜0或0＜t＜2

liman(n→∞）=2limbn(n→∞）=2/(t-2).

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