已知an是等差数列且a1=3,a1+a2+a3=15,(1)求数列an的通项（2）求数列1/an*an+1的前N项和Sn

Question

已知an是等差数列 且a1=3,a1+a2+a3=15, (1)求数列an的通项 （2）求数列1/an*an+1的前N项和Sn

Answer 1

1、
等差则a1+a3=2a2
所以a1+a2+a3=3a2=15
a2=5
a1=3
所以d=a2-a1=2
所以an=2n+1
2、
1/ana(n+1)
=1/(2n+1)(2n+3)
=1/2*2/(2n+1)(2n+3)
=(1/2)*[(2n+3)-(2n+1)]/(2n+1)(2n+3)
=(1/2)*[(2n+3)/(2n+1)(2n+3)-(2n+1)/(2n+1)(2n+3)]
=(1/2)*[1/(2n+1)-1/(2n+3)]
所以Sn=(1/2)*[1/3-1/5+1/5-1/7+……+1/(2n+1)-1/(2n+3)]
=(1/2)*[1/3-1/(2n+3)]
=n/(6n+9)