Question

已知等差数列【an】满足:a3=7,a5+a7=26.【an】的前n项和为Sn.(1)求a4及Sn;

(2)bn=1/(an^2-1)(n属于N*),求数列【bn】的前n项和Tn

Answer 1

(1)因为a5+a7=26
所以a6=13则公差d=(13-7)/(6-3)=2
所以an=7+2(n-3)=2n+1
所以a4=9.Sn=(2n+1)(n+1)
(2)bn=1/2n(2n+2)=1/4(1/n-1/(n+1))
所以Tn=1/4(1-1/2+1/2-2/3+2/3-3/4+...+1/n-1/(n+1))=1/4*n/(n+1)=n/4(n+1)

Answer 2

设等差为d，则a5 = a3+ 2d，a7 = a3+ 4d 故有：
a3 + 2d + （a3 + 4d）=26
把a3 = 7代入上式得：d = 2
所以 a4 = a3 + d = 9
a1 = a3 - 2d = 3
所以 Sn = na1 + [n(n-1)d]/2
= 3n + n^2 - n = n^2 + 2n

Answer 3

a5=a3+2d，a7=a3+4d
a5+a7=2a3+6d
26=2*7+6d
d=2
a4=a3+d=7+2=9
a1=a3-2d=7-2*2=3
Sn=na1+n(n-1)d/2=n^2+2n

an=a1+nd=3+2n
bn=1/(an^2-1)=1/[4(n+2)(n+1)]=1/4[(1/(n+1)-1/(n+2)]