已知数列{an}的前n项和为Sn,且a1=2,Sn=2an+k,等差数列{bn}的前n项和为Tn,且Tn=n2...
问题详情:
已知数列{an}的前n项和为Sn,且a1=2,Sn=2an+k,等差数列{bn}的前n项和为Tn,且Tn=n2.
(1)求k和Sn;
(2)若cn=an·bn,求数列{cn}的前n项和Mn.
【回答】
解 (1)∵Sn=2an+k,
∴当n=1时,S1=2a1+k.
∴a1=-k=2,即k=-2.
∴Sn=2an-2.
∴当n≥2时,Sn-1=2an-1-2.
∴an=Sn-Sn-1=2an-2an-1.
∴an=2an-1.
∴数列{an}是以2为首项,2为公比的等比数列.∴{an}=2n.
∴Sn=2n+1-2.
(2)∵等差数列{bn}的前n项和为Tn,且Tn=n2,
∴当n≥2时,bn=Tn-Tn-1=n2-(n-1)2=2n-1.
又b1=T1=1符合bn=2n-1,
∴bn=2n-1.∴cn=an·bn=(2n-1)2n.
∴数列{cn}的前n项和
Mn=1×2+3×22+5×23+…+(2n-3)×2n-1+(2n-1)×2n,①
∴2Mn=1×22+3×23+5×24+…+(2n-3)×2n+(2n-1)×2n+1,②
由①-②,得-Mn=2+2×22+2×23+2×24+…+2×2n-(2n-1)×2n+1
=2+2×-(2n-1)×2n+1,
即Mn=6+(2n-3)2n+1.
知识点:数列
题型:解答题