等比数列{an}的前n项和为Sn,Sn=2 的n次方+m,(1)求m(2)bn=4an分之n+1,求bn前n项和
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/07/06 02:30:18
等比数列{an}的前n项和为Sn,Sn=2 的n次方+m,(1)求m(2)bn=4an分之n+1,求bn前n项和
等比数列{an}的前n项和为Sn,Sn=2
的n次方+m,(1)求m(2)bn=4an分之n+1,求bn前n项和
等比数列{an}的前n项和为Sn,Sn=2
的n次方+m,(1)求m(2)bn=4an分之n+1,求bn前n项和
设公比为q,q=1时,Sn=na1≠2ⁿ+m,因此q≠1
Sn=a1(qⁿ-1)/(q-1)=[a1/(q-1)]qⁿ -[a1/(q-1)]
Sn=2ⁿ+m
q=2
a1/(q-1)=1
m=-a1/(q-1)
解得m=-1 a1=1 q=2
m=-1
an=a1q^(n-1)=1×2^(n-1)=2^(n-1)
bn=(n+1)/4an=(n+1)/2^(n+1)
Tn=b1+b2+...+bn
=2/2²+3/2³+4/2⁴+...+(n+1)/2^(n+1)
Tn /2=2/2³+3/2⁴+...+n/2^(n+1) +(n+1)/2^(n+2)
Tn- Tn /2=Tn /2=2/2²+1/2³+1/2⁴+...+1/2^(n+1) -(n+1)/2^(n+2)
=1/2+1/2²+...+1/2^(n+1) -(n+1)/2^(n+2) -1/4
=(1/2)×[1-(1/2)^(n+1)]/(1-1/2) -(n+1)/2^(n+2) -1/4
=3/4- (n+3)/2^(n+2)
Tn=3/2 -(n+3)/2^(n+1)
Sn=a1(qⁿ-1)/(q-1)=[a1/(q-1)]qⁿ -[a1/(q-1)]
Sn=2ⁿ+m
q=2
a1/(q-1)=1
m=-a1/(q-1)
解得m=-1 a1=1 q=2
m=-1
an=a1q^(n-1)=1×2^(n-1)=2^(n-1)
bn=(n+1)/4an=(n+1)/2^(n+1)
Tn=b1+b2+...+bn
=2/2²+3/2³+4/2⁴+...+(n+1)/2^(n+1)
Tn /2=2/2³+3/2⁴+...+n/2^(n+1) +(n+1)/2^(n+2)
Tn- Tn /2=Tn /2=2/2²+1/2³+1/2⁴+...+1/2^(n+1) -(n+1)/2^(n+2)
=1/2+1/2²+...+1/2^(n+1) -(n+1)/2^(n+2) -1/4
=(1/2)×[1-(1/2)^(n+1)]/(1-1/2) -(n+1)/2^(n+2) -1/4
=3/4- (n+3)/2^(n+2)
Tn=3/2 -(n+3)/2^(n+1)
数列{an}的前n项和为Sn=n平方+n,(1)求an,(2)令bn=2的an次方,证明bn为等比数列,并求前n项和Tn
1.等差数列{an},{bn}的前n项和分别为Sn,Tn,(1)若Sm=n,Sn=m,求Sn+m
数列an的前n项和为Sn,Sn=4an-3,①证明an是等比数列②数列bn满足b1=2,bn+1=an+bn.求数列bn
数列An=a的n次方,Sn为An的前n项和,若Bn=2Sn/An+1,且bn为等比数列,求a的值
设等差数列{an}的前 n项和为Sn,且 Sn=(an+1)^/2(n属于N*)若bn=(-1)nSn,求数列{bn}的
等差数列an,bn的前n项和分别为Sn,若Sn/Tn=2n/(3n+1),求an/bn的表达式
等差数列{an},{bn}的前n项和分别为Sn,Tn,若Sn/Tn=2n/3n+1,求an/bn的表达式
等差数列{an}、{bn}的前n项和分别为Sn、Tn,若Sn/Tn=2n/3n+1,求an/bn
两个等差数列{an},{bn}的前n项和分别为Sn,Tn,若Sn/Tn=2n/3n+1,求an/bn.
等差数列an,bn的前n项和分别为Sn,若Sn/Tn=2n/(3n+1),求lim an/bn
an=2的n-1次方,bn=2n-1,设数列an的前n项和为sn,求数列{sn.bn}的前n项和Tn
数列{bn}的前n项和为Sn,且Sn,且Sn=1-1/2bn(n∈N+) 求{bn}的通项公式