设数列{an}的前n项和为Sn,若SnS2n为常数,则称数列{an}为“吉祥数列“,己知等差数列{bn}的首项为1,公差不为0,若数列{bn}为“吉祥数列“,则数列{bn}的通项公式为()A.bn=n-1B.bn=2n-1C.
设数列{an}的前n项和为Sn,若
为常数,则称数列{an}为“吉祥数列“,己知等差数列{bn}的首项为1,公差不为0,若数列{bn}为“吉祥数列“,则数列{bn}的通项公式为( )Sn S2n
A. bn=n-1
B. bn=2n-1
C. bn=n+1
D. bn=2n+1
由
Sn |
S2n |
得n+
1 |
2 |
1 |
2 |
即2+(n-1)d=4k+2k(2n-1)d.
整理得,(4k-1)dn+(2k-1)(2-d)=0.
∵对任意正整数n上式恒成立,
则
|
|
∴数列{bn}的公差为2,
则其通项公式为bn=1+2(n-1)=2n-1,
故选:B.
已知f(x)=x/(1+x),数列{an}为首项是1,以f(1)为公比的等比数列;数列{bn}中b 2020-05-13 …
已知数列{an}和{Bn}满足a1=2 an-1=an(an+1-1) bn=an-1 n∈N+已 2020-05-15 …
设数列an的前n项和为sn,点P(Sn,an)在直线(3-m)x+2my-m-3=0上,m属于N* 2020-06-05 …
bn+1=2bn²-bn+1/2求通项!我要的是方法!公式!已知f(x)=x²-1/2x+1/4若 2020-07-09 …
若等差数列{an}中,a1=3,a4=12,{bn-an}为等比数列,且数列{bn}满足b1=4, 2020-07-09 …
已知数列{an}的前n项和为Sn,且Sn=n^2+11n,数列{bn}满足b(n+2)-2b(n+ 2020-07-18 …
对于无穷数列{an}与{bn},记A={x|x=an,n∈N*},B={x|x=bn,n∈N*}, 2020-07-22 …
对数列{an}和{bn},若对任意正整数n,恒有bn≤an,则称数列{bn}是数列{an}的“下界 2020-07-31 …
(2012•静安区一模)下列命题中正确的命题是()A.若limn→∞an=A,limn→∞bn=B, 2020-12-16 …
数列{an}的通项公式an=2的n-1次方,数列{bn}是等差数列,令集合A={a1,a2,…an, 2021-02-09 …