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已知数列{an}的首项a1=1,且满足an+1=an4an+1(n∈N*).(Ⅰ)设bn=1an,求证:数列{bn}是等差数列,并求数列{an}的通项公式;(Ⅱ)设cn=bn•2n,求数列{cn}的前n项和Sn.
题目详情
已知数列{an}的首项a1=1,且满足an+1=
(n∈N*).
(Ⅰ)设bn=
,求证:数列{bn}是等差数列,并求数列{an}的通项公式;
(Ⅱ)设cn=bn•2n,求数列{cn}的前n项和Sn.
an |
4an+1 |
(Ⅰ)设bn=
1 |
an |
(Ⅱ)设cn=bn•2n,求数列{cn}的前n项和Sn.
▼优质解答
答案和解析
(Ⅰ)∵an+1=
,∴
=4+
,即
−
=4,∴bn+1-bn=4.
∴数列{bn}是以1为首项,4为公差的等差数列.∴
=bn=1+4(n−1)=4n-3,
∴数列{an}的通项公式为an=
.
(Ⅱ)由(Ⅰ)知cn=bn•2n=(4n-3)2n,
∴Sn=1×21+5×22+9×23+…+(4n−3)•2n…①
同乘以2得,2Sn=1×22+5×23+9×24+…+(4n−3)•2n+1…②
②-①得,−Sn=2−(4n−3)2n+1+4×(22+23+24+…+2n)
=2−(4n−3)2n+1+
=2-(4n-3)2n+1+16×2n-1-16
=-14+2n+1×(4-4n+3)=-14+(7-4n)2n+1
∴数列{cn}的前n项和Sn=(4n−7)•2n+1+14
an |
4an+1 |
1 |
an+1 |
1 |
an |
1 |
an+1 |
1 |
an |
∴数列{bn}是以1为首项,4为公差的等差数列.∴
1 |
an |
∴数列{an}的通项公式为an=
1 |
4n−3 |
(Ⅱ)由(Ⅰ)知cn=bn•2n=(4n-3)2n,
∴Sn=1×21+5×22+9×23+…+(4n−3)•2n…①
同乘以2得,2Sn=1×22+5×23+9×24+…+(4n−3)•2n+1…②
②-①得,−Sn=2−(4n−3)2n+1+4×(22+23+24+…+2n)
=2−(4n−3)2n+1+
4×22×(1−2n−1) |
1−2 |
=-14+2n+1×(4-4n+3)=-14+(7-4n)2n+1
∴数列{cn}的前n项和Sn=(4n−7)•2n+1+14
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