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
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/10/07 01:56:38
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→∞ |
(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.
已知数列an为等差数列,公差d≠0,bn为等比数列,若b1=a1,b2=a3,b3=a2,公比q=?
数学数列难题已知等差数列{an}公差为d,d不等于0,等比数列{bn}公比q,q大于1.设Sn =a1b1 +a2b2
已知数列[an]为等差数列,公差d≠0;[bn]为等比数列,公比为q,若a1=b1,a3=b3,a7=b5,且an=bm
已知数列an为等差数列,公差d≠0,bn为等比数列,公比为q,
在公差为d的等差数列an和公比q的等比数列bn中,a2=b1=3,a5=b2,a14=b3.求anbn通项公式拜托各位了
已知f(x)=(x-1)^2,数列{an}是公差为d的等差数列,{bn}是公比为q的等比数列,设a1=f(d-1),a3
已知f(x)=(x-1)^2,数列{an}是公差为d的等差数列,{bn}是公比为q的等比数列,设a1=f(d-1)
设等差数列{an}的公差d≠0,数列{bn}为等比数列,若a1=b1,b2=a3 b3=a2,则bn的公比为
已知数列{an}是首项为1公差为正的等差数列,数列{bn}是首项为1的等比数列,设Cn=anbn(n∈N*),且数列{c
1``已知等差数列{An}的公差和等比数列{Bn}的公比都是d(d不为0)且a1=b1,a4=b4,a10=b10.
在公差为d(d≠0)的等差数列{an}和公比为q的等比数列{bn}中,已知a1=b1=1,a2=b2,a8=b3,求d
已知数列an为等差数列,公差d≠0,bn为等比数列,若b1=a1,b2=a3,b3=a2,公比是多少