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an及Sn; (2)令bn=1/an^2-1(n属于N),求数...

已知等差数列{an}满足:a3=7,a5+a7=26.{an}的前n项和为Sn.
(1)求an及Sn;
(2)令bn=1/an^2-1(n属于N),求数列{bn}的前n项和Tn.
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Answer 1

1)a5+a7=2a6=26得a6=13
d=(a6-a3)/(6-3)=(13-7)/3=2
a1=a3-2d=3
Sn=3n+(1/2)n(n-1)·2=n^2+2n
2)an=2n+1,然后写出bn的式子,利用拆项求和,得Tn=n/[4(n+1)]

Answer 2

1.由a5+a7=26得a6=13,故公差d=(a6-a3)/(6-3)=2,所以an=a3+(n-3)d=2n+1,sn=n(n+2)
2.bn=1/(an^2-1)=1/4n(n+1)=(1/4)[1/n -1/(n+1)]
故Tn=(1/4)[1-1/(n+1)]=n/[4(n+1)]

Answer 3

a6=13，a3=7,d=2.a1=3,所以an=3+(n-1)*2=2n+1，sn=n*a1+n(n-1)d/2=n^2+2n
bn=1/an^2-1,bn=1/(2n+1)^2-1=1/4(1/n-1/(n+1))
可知b1=1/4(1-1/2),
Tn=1/4(1-1/2+…1/n-1/(n+1))=1/4(1-1/(n+1))=n/4(n+1)

Answer 4

a5+a7=2a6=26，得到a6=13。a6-a3=3d=13-7=6，得到d=2。.
可得a3-2d=a1=7-4=3。.
Sn=na1+n(n-1)d/2=3n+n(n-1)=n^2+2n。

an=3+2(n-1)=2n+1.

bn=1/(2n+1)^2-1=-(4n^2+4n)/(2n+2)^2=-n(n+1)=1/(n+1)-1/n。

tn=b1+b2+b3+……+bn=1/2-1+1/3-1/2+1/4-1/3+…………+1/(1+n)-1/n=1/(1+n)-1