Question

在等差数列an中，A7=4 A19=2a9求数列an的通项公式 若bn=1/nan 求数

在等差数列an中，A7=4 A19=2a9求数列an的通项公式 若bn=1/nan 求数列bn的前n项和Sn

Answer 1

a9=a7+2d=4+2d
a19=a7+12d=4+12d
得
4+12d=2*(4+2d)
即d=1/2
a1=a4-3d
=4-3/2
=5/2
an=a1+(n-1)d
=5/2+3n/2-3/2
=(3n+2)/2

追问

第二题

追答

a9=a7+2d=4+2d
a19=a7+12d=4+12d
得
4+12d=2*(4+2d)
即d=1/2
a1=a7-6d
=4-3
=1
an=a1+(n-1)d
=1+n/2-1/2
=(n+1)/2

bn=1/nan
=1+1/(2*3/2)+1/(3*2)..........1/n*(n+1)/2
=1+1/3+1/6+1/10+....1/n*(n+1)/2
=2*(1/2+1/6+1/12+1/20+.......1/n*(n+1)）
=2*((1-1/2)+(1/2-1/3)+(1/3-1/4)+(1/4-1/5)+....(1/n-1/(n+1))
=2*(1-1/(n+1))
=2*n/(n+1)
= 2n/(n+1)

因为bn=1/nan
所以
Sbn=1+1/(2*3/2)+1/(3*2)..........1/n*(n+1)/2
=1+1/3+1/6+1/10+....1/n*(n+1)/2
=2*(1/2+1/6+1/12+1/20+.......1/n*(n+1)）
=2*((1-1/2)+(1/2-1/3)+(1/3-1/4)+(1/4-1/5)+....(1/n-1/(n+1))
=2*(1-1/(n+1))
=2*n/(n+1)
= 2n/(n+1)