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/
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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
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
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
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