已知数列{an}是公差不为0的等差数列,a1=1,且a2,a4,a8成等比数列....
解答:解:(1)∵数列{an}是公差不为0的等差数列,a1=1,且a2,a4,a8成等比数列,∴(1+3d)2=(1+d)(1+7d),
解得d=1,或d=0(舍),
∴an=1+(n-1)×1=n.
(2)设{bn}的公比为q,
∵
Sn
2
=15,
S2n
2
=255,
∴S2n=
b1(1-q2n)
1-q
=510,Sn=
b1(1-qn)
1-q
30,
两式相除,得1+qn=17,
∴bn=b1qn-1=
b1
q
•qn16•
b1
q
,
∵在前n项和中,最大项为16,
∴只有
b1
q
=1时最大,故b1=q时取得.
将所得结果代入到
Sn
2
=15,求得b1=q=2,bn=2n.
cn=an•bn=n•2n,
Tn=1•2+2•22+3•23+…+n•2n,①
2Tn=1•22+2•23+3•24+…+n•2n+1,②
①-②,得-Tn=2+22+23+…+2n-n•2n+1
=2n+1-2-n•2n+1
=-(n-1)•2n+1-2,
∴Tn=(n-1)•2n+1+2.
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