a1=m2/4an为等比数列,q=m/2bn=anSn/n2Tn=b1+b2```````bn比较Tn与a1/2的大小
来源:学生作业帮 编辑:搜狗做题网作业帮 分类:数学作业 时间:2024/07/02 19:43:32
a1=m2/4
an为等比数列,q=m/2
bn=anSn/n2
Tn=b1+b2```````bn
比较Tn与a1/2的大小
an为等比数列,q=m/2
bn=anSn/n2
Tn=b1+b2```````bn
比较Tn与a1/2的大小
![a1=m2/4an为等比数列,q=m/2bn=anSn/n2Tn=b1+b2```````bn比较Tn与a1/2的大小](/uploads/image/z/17244111-39-1.jpg?t=a1%3Dm2%2F4an%E4%B8%BA%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%2Cq%3Dm%2F2bn%3DanSn%2Fn2Tn%3Db1%2Bb2%60%60%60%60%60%60%60bn%E6%AF%94%E8%BE%83Tn%E4%B8%8Ea1%2F2%E7%9A%84%E5%A4%A7%E5%B0%8F)
a1=m2/4
an为等比数列,q=m/2
an=a1*q^(n-1)=m^2/4*(m/2)^(n-1)=m^(n+1)/2^(n+1)
Sn=a1*(q^n-1)/(q-1)=m^2/4*(m^2/4-1)/(m/2-1)=m^2/4*(m^2-4)/(2m-4)=m^2/4*(m+2)/2=m^2(m+2)/8
bn=an*Sn/n^2=m^(n+1)/2^(n+1)*m^2(m+2)/8*1/n^2
=m^(n+3)*(m+2)/[2^(n+4)*n^2]
an为等比数列,q=m/2
an=a1*q^(n-1)=m^2/4*(m/2)^(n-1)=m^(n+1)/2^(n+1)
Sn=a1*(q^n-1)/(q-1)=m^2/4*(m^2/4-1)/(m/2-1)=m^2/4*(m^2-4)/(2m-4)=m^2/4*(m+2)/2=m^2(m+2)/8
bn=an*Sn/n^2=m^(n+1)/2^(n+1)*m^2(m+2)/8*1/n^2
=m^(n+3)*(m+2)/[2^(n+4)*n^2]
an=2^n bn=2n Tm=b1/a1+b2/a2+……+bn/an,求Tn
设{an}为等差数列,且等比数列{bn}中有b1=a1^2,b2=a2^2,b3^2(a1
数列{an}的前n项和为sn=2n^2,{bn}为等比数列,且a1=b1,b2(a2-a1)=b1
设数列{an}的前n项和为Sn=2n^2,{bn}为等比数列,且a1=b1,b2(a2-a1)=b1.
设数列{an}的前n项和为Sn=2n²,{bn}为等比数列,且a1=b1,b2(a2-a1)=b1,
设数列{an}的前n项和为Sn=2n2,{bn}为等比数列,且a1=b1,b2(a2-a1)=b1.
设数列{an}的前n项和为Sn=2n平方,{bn}为等比数列,且a1=b1,b2(a2-a1)=b1
设数列{an}的前n项和为Sn=2n²{bn}为等比数列,且a1=b1,b2(a2-a1)=b1
设数列{an}的前n项和胃Sn=2n^2,{bn}为等比数列,且a1=b1,b2(a2-a1)=b1
设数列{An}的前n项伟Sn=2n^2,{Bn}为等比数列,且a1=b1,(a2-a1)b2=b1
已知数列{an}是等差数列,且a1=1,公差为2,数列{bn}为等比数列且b1=a1,b2(a2-a1)=b1
已知数列{an}是等差数列,a1=1,公差为2,又已知数列{bn}为等比数列,且b1=a1,b2(a2-a1)=b1,求