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
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![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/2180663-71-3.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=m2/4an为等比数列,q=m/2bn=anSn/n2Tn=b1+b2```````bn比较Tn与a1/2的大小
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
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
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]
用代入法把
Tn大,你想嘛。
顶楼上的,接下来可以用列举法。比较T1,T2,T3与之的大小。然后就可以下结论了