Question

高一等差等比数列的证明

1.数列{an}满足an=1,
若bn等于an-2，求bn为等比数列
求an通项
2.数列｛an｝中，a1=2,a(n+1)=3/2an+1,求an
呃呃呃，是a1=1,不好意思啊

Answer 1

1.设bn等比为q,
b1=a1-2=-1
bn=-q^(n-1)
==>an=bn+2=2-q^(n-1)
2.a(n+1)=3/2an+1
令a(n+1)+t=3/2(an+t)
==>a(n+1)=3/2an+t/2
==>t=2
==>a(n+1)+2=3/2(an+2)
令bn=an+2,b1=a1+2=4
==>b(n+1)=3/2bn
==>bn是以4为首项，公比为3/2的等比数列
==>bn=4*(3/2)^(n-1)
==>an=bn-2=4*(3/2)^(n-1)-2

Answer 2

an=1?那就是常数列了好不 题目都写不清楚人家怎么回答啊