设(an)为公差大于0的等差数列,sn为其前n项和,s4=24,a2a3=35,（1）求(an)通项公式（2）若bn=1/an*a(n+1),求(bn)前n项和Tn.

Question

设(an)为公差大于0的等差数列,sn为其前n项和,s4=24,a2a3=35,（1）求(an)通项公式
（2）若bn=1/an*a(n+1),求(bn)前n项和Tn.

Answer 1

24
=S4
=a1+a2+a3+a4
=2(a2+a3)
=>
a2+a3=12
a2*a3=35
=>
a2=5,a3=7
=>
a1=3
=>
an=3+(n-1)*2=2n+1
bn
=1/an*a(n+1)
=1/((2n+1)(2n+3))
=1/2*(1/(2n+1)-1/(2n+3))
=>
Tn
=0.5*(1/3-1/5+1/5-1/7+...+1/(2n+1)-1/(2n+3))
=0.5*(1/3-1/(2n+3))