Question

已知正项数列an满足an2-nan-(n+1)=0，数列bn的前n项和为Sn，且Sn=2bn-2(1

已知正项数列an满足an2-nan-(n+1)=0，数列bn的前n项和为Sn，且Sn=2bn-2(1)求数列an，bn的通项公式
(2)求数列1/(anlog2bn)的前n项和Tn

Answer 1

an^2-nan-(n+1)=0
(an +1)[an-(n+1)]=0
an=-1(数列为正项数列，an>0，舍去)或an=n+1
n=1时，b1=S1=2b1-2
b1=2
n≥2时，
bn=Sn-S(n-1)=2bn -2-[2b(n-1)-2]
bn=2b(n-1)
bn/b(n-1)=2，为定值，数列{bn}是以2为首项，2为公比的等比数列
bn=2×2^(n-1)=2ⁿ
数列{an}的通项公式为an=n+1；数列{bn}的通项公式为bn=2ⁿ
1/[anlo2(bn)]
=1/[(n+1)log2(2ⁿ)]
=1/[n(n+1)]
=1/n -1/(n+1)
Tn=1/1-1/2+1/2-1/3+...+1/n -1/(n+1)
=1-1/(n+1)
=n/(n+1)