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
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![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](/uploads/image/z/935065-1-5.jpg?t=%E5%B7%B2%E7%9F%A5%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%E3%80%90an%E3%80%91%E6%BB%A1%E8%B6%B3%3Aa3%3D7%2Ca5%2Ba7%3D26.%E3%80%90an%E3%80%91%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn.%281%29%E6%B1%82a4%E5%8F%8ASn%3B%282%29bn%3D1%2F%28an%5E2-1%29%28n%E5%B1%9E%E4%BA%8EN%2A%29%2C%E6%B1%82%E6%95%B0%E5%88%97%E3%80%90bn%E3%80%91%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8CTn)
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
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
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
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)]
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
(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)