已知等差数列{an}的前n项和为Sn,且a1=1,S9=81,(1)求数列{an}的通项公式,(2)令bn=1/(an乘a(n+1)),数列{bn}的前n项和为Tn,求Tn的取值范围

Question

已知等差数列{an}的前n项和为Sn,且a1=1,S9=81,(1)求数列{an}的通项公式,(2)令bn=1/(an乘a(n+1)),数列{bn}
的前n项和为Tn,求Tn的取值范围

Answer 1

S9=9a1+(1+8)*8/2*k=9+36k=81k=2an=1+2(n-1)bn=1/[an*a(n+1)]=1/(2n-1)(2n+1) Tn=1/*3+1/3*5+……+1/（2n-1)(2n+1)=1/2(1-1/3+1/3-1/5+……+1/(2n-1)-1/(2n+1)]1/2[1-1/(2n+1)]=n/(2n+1)1/3