已知数列{an}满足a1=2,an+1=2an+3.
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已知数列{an}满足a1=2,an+1=2an+3.
(1)求{an}的通项公式;
(2)求数列{nan}的前n项和Sn.
(1)求{an}的通项公式;
(2)求数列{nan}的前n项和Sn.
(1)∵a1=2,an+1=2an+3.
∴an+1+3=2(an+3),a1+3=5
∴数列{an+3}是以5为首项,以2为公比的等比数列
∴an+3=5•2n−1
∴an=5•2n−1−3
(2)∵nan=5n•2n−1−3n
令Tn=1•20+2•21+…+n•2n−1
则2Tn=1•2+2•22+…+(n-1)•2n-1+n•2n
两式相减可得,-Tn=1+2+22+…+2n-1-n•2n=
1−2n
1−2-n•2n=2n-n•2n-1
∴Tn=(n−1)•2n+1
∴Sn=5(n−1)•2n−
3n2+3n
2+5
∴an+1+3=2(an+3),a1+3=5
∴数列{an+3}是以5为首项,以2为公比的等比数列
∴an+3=5•2n−1
∴an=5•2n−1−3
(2)∵nan=5n•2n−1−3n
令Tn=1•20+2•21+…+n•2n−1
则2Tn=1•2+2•22+…+(n-1)•2n-1+n•2n
两式相减可得,-Tn=1+2+22+…+2n-1-n•2n=
1−2n
1−2-n•2n=2n-n•2n-1
∴Tn=(n−1)•2n+1
∴Sn=5(n−1)•2n−
3n2+3n
2+5
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