设数列{an}、{bn}满足:a1=b1=6,a2=b2=4,a3=b3=3,且数列{an+1-an}是等差数列,{bn-2}是等比数列,其中n∈N*. (1)求数列{an}和{bn}的通项公式; (2)求数列{bn}的前n项和Sn.
问题描述:
设数列{an}、{bn}满足:a1=b1=6,a2=b2=4,a3=b3=3,且数列{an+1-an}是等差数列,{bn-2}是等比数列,其中n∈N*.
(1)求数列{an}和{bn}的通项公式;
(2)求数列{bn}的前n项和Sn.
答
(1)因为 {an+1-an}是等差数列,所以a2-a1=-2,a3-a2=-1,a4-a3=0,…,an-an-1=n-4,以上各式相加得,an-a1=(n−1)(n−6)2,即an=6+(n−1)(n−6)2(n≥2),又a1=6,所以an=6+(n−1)(n−6)2;b1-2=4,b2-2=2,所以...