Question

已知等差数列｛An｝的前n项和为Sn，且A3=7，S11=143

已知等差数列｛An｝的前n项和为Sn，且A3=7，S11=143。（1）求数列｛An｝的通项公式An，（2）令Bn=4/An*An-1(n∈N*)，求数列{Bn}的前n项和Tn

Answer 1

解：
Sn=（A1+An）*n/2
S11=（A1+A11）*11/2
=（A3-2d+A3+8d）*11/2
=（A3+3d）*11
=（7+3d）*11
=143
所以，d=2，A1=A3-2d=7-4=3
通项公式An=A1+（n-1）d
=3+2（n-1）
=2n+1
Bn=4/(An^2-1)
=4/［(2n+1)^2-1］
=4/(4n^2+4n+1-1)
=1/(n^2+n)
=1/［n*(n+1)］
=1/n-1/(n+1)
Tn=B1+B2+B3+……+Bn
=1/1-1/2+1/2-1/3+1/3-1/4+……+1/（n-1）-1/n
=1-1/n