设数列{an}满足a1=a,an+1=can +1-c,其中a,c为实数,且c≠0 求:若0

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设数列{an}满足a1=a,an+1=can +1-c,其中a,c为实数,且c≠0 求:若0

设数列{an}满足a1=a,an+1=can +1-c,其中a,c为实数,且c≠0 求:若0
设数列{an}满足a1=a,an+1=can +1-c,其中a,c为实数,且c≠0 求:若0

设数列{an}满足a1=a,an+1=can +1-c,其中a,c为实数,且c≠0 求:若0
由a(n+1)=can+1-c,移项:a(n+1)-1=c(an-1),得an-1=c^n*(a0-1);
an=1+c^n*(a-1)(n>=0)…………(1);
若a=1,则an=1(n>=0),这与01,矛盾;
因此a1,由于c^n->+∞(n->+∞),当n充分大时,an

A(n+1)-1=c(An-1) 故0

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