已知Sn为数列{an}的前n项和,Sn=12n2+112n;数列{bn}满足:b3=11,bn+2=2bn+1-bn,其
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已知Sn为数列{an}的前n项和,Sn=
n
1 |
2 |
(I)∵数列{an}的前n项和,Sn=
1
2n2+
11
2n.
∴当n=1时,a1=S1=
1
2+
11
2=6;
当n≥2时,an=Sn-Sn-1=
1
2n2+
11
2n-[
1
2(n−1)2+
11
2(n−1)]=n+5.
当n=1时,上式成立,
∴an=n+5.
∵b3=11,bn+2=2bn+1-bn,
∴数列{bn}是等差数列,设公差为d.
∵前9项和为153,
∴153=9b1+
9×8
2d,b3=b1+2d=11.解得b1=5,d=3.
∴bn=5+3(n-1)=3n+2.
(II)cn=
6
(2an−11)(2bn−1)=
6
(2n+10−11)(6n+4−1)=
1
2n−1−
1
2n+1,
∴Tn=(1−
1
3)+(
1
3−
1
5)+…+(
1
2n−1−
1
2n+1)
=1−
1
2n+1
=
2n
2n+1.
1
2n2+
11
2n.
∴当n=1时,a1=S1=
1
2+
11
2=6;
当n≥2时,an=Sn-Sn-1=
1
2n2+
11
2n-[
1
2(n−1)2+
11
2(n−1)]=n+5.
当n=1时,上式成立,
∴an=n+5.
∵b3=11,bn+2=2bn+1-bn,
∴数列{bn}是等差数列,设公差为d.
∵前9项和为153,
∴153=9b1+
9×8
2d,b3=b1+2d=11.解得b1=5,d=3.
∴bn=5+3(n-1)=3n+2.
(II)cn=
6
(2an−11)(2bn−1)=
6
(2n+10−11)(6n+4−1)=
1
2n−1−
1
2n+1,
∴Tn=(1−
1
3)+(
1
3−
1
5)+…+(
1
2n−1−
1
2n+1)
=1−
1
2n+1
=
2n
2n+1.
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