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
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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.
(1)求数列{an}和{bn}的通项公式;
(2)求数列{bn}的前n项和Sn.
(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,所以公比为
1
2,
所以bn−2=4•(
1
2)n−1=23-n,故bn=23−n+2;
(2)Sn=b1+b2+b3+…+bn=2n+
4[1−(
1
2)n]
1−
1
2=2n+8-23-n.
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,所以公比为
1
2,
所以bn−2=4•(
1
2)n−1=23-n,故bn=23−n+2;
(2)Sn=b1+b2+b3+…+bn=2n+
4[1−(
1
2)n]
1−
1
2=2n+8-23-n.
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