已知数列{an}为等差数列,公差为d,{bn}为等比数列,公比为q,且d=q=2,b3+1=a10=5,设cn=anbn
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已知数列{an}为等差数列,公差为d,{bn}为等比数列,公比为q,且d=q=2,b3+1=a10=5,设cn=anbn.
(1)求数列{cn}的通项公式;
(2)设数列{cn}的前n项和为Sn,
(3)(理)求
(1)求数列{cn}的通项公式;
(2)设数列{cn}的前n项和为Sn,
(3)(理)求
lim |
n→∞ |
![已知数列{an}为等差数列,公差为d,{bn}为等比数列,公比为q,且d=q=2,b3+1=a10=5,设cn=anbn](/uploads/image/z/15462374-14-4.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%E4%B8%BA%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%EF%BC%8C%E5%85%AC%E5%B7%AE%E4%B8%BAd%EF%BC%8C%7Bbn%7D%E4%B8%BA%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%EF%BC%8C%E5%85%AC%E6%AF%94%E4%B8%BAq%EF%BC%8C%E4%B8%94d%3Dq%3D2%EF%BC%8Cb3%2B1%3Da10%3D5%EF%BC%8C%E8%AE%BEcn%3Danbn)
(1)∵a10=5,d=2,∴an=2n-15.
又∵b3=4,q=2,∴bn=2n-1,∴cn=(2n-15)•2n-1.
(2)∵Sn=c1+c2+c3+…+cn,∴2Sn=2c1+2c2+2c3+…+2cn,
错位相减,得-Sn=c1+(c2-2c1)+(c3-2c2)+…+(cn-2cn-1)-2cn.
∵c1=-13,cn-2cn-1=2n,
∴-Sn=-13+22+23+…+2n-(2n-15)•2n=-13+4(2n-1-1)-(2n-15)•2n
=-17+2n+1-(2n-15)•2n∴Sn=17+(2n-17)•2n.
∴
lim
n→∞
nbn
Sn=
lim
n→∞
n•2n−1
17+(2n−17)•2n
=
lim
n→∞
n
17
2n−1+(2n−17)•2=
1
4.
又∵b3=4,q=2,∴bn=2n-1,∴cn=(2n-15)•2n-1.
(2)∵Sn=c1+c2+c3+…+cn,∴2Sn=2c1+2c2+2c3+…+2cn,
错位相减,得-Sn=c1+(c2-2c1)+(c3-2c2)+…+(cn-2cn-1)-2cn.
∵c1=-13,cn-2cn-1=2n,
∴-Sn=-13+22+23+…+2n-(2n-15)•2n=-13+4(2n-1-1)-(2n-15)•2n
=-17+2n+1-(2n-15)•2n∴Sn=17+(2n-17)•2n.
∴
lim
n→∞
nbn
Sn=
lim
n→∞
n•2n−1
17+(2n−17)•2n
=
lim
n→∞
n
17
2n−1+(2n−17)•2=
1
4.
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