已知等差数列{an}满足:a3=7,a5+a7=26.{an}的前n项和为Sn.求令bn=1/(an)^2-1,求{bn
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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.求令bn=1/(an)^2-1,求{bn}及前n项和Tn
bn是1/[(an)^2-1]
bn是1/[(an)^2-1]
a3=7
a5+a7=2a6=26
a6=13
a6-a3=6=5d-2d=3d,d=2
a1+2d=7=a1+4 a1=3
an=3+2(n-1)=2n+1
bn=1/[an^2-1]=1/[4n(n+1)]=(1/4)(1/n-1/(n+1))
b1=(1/4)(1-1/2)=1/8
Tn=(1/4)(1-1/(n+1))
再问: Tn=(1/4)(1-1/(n+1)) 是什么啊?
再答: b1= (1/4)(1-1/2) b2=(1/4)(1/2-1/3) .. bn=(1/4)(1/n-1/(n+1)) 左和 =右和 Sn=(1/4)(1-1/(n+1))
a5+a7=2a6=26
a6=13
a6-a3=6=5d-2d=3d,d=2
a1+2d=7=a1+4 a1=3
an=3+2(n-1)=2n+1
bn=1/[an^2-1]=1/[4n(n+1)]=(1/4)(1/n-1/(n+1))
b1=(1/4)(1-1/2)=1/8
Tn=(1/4)(1-1/(n+1))
再问: Tn=(1/4)(1-1/(n+1)) 是什么啊?
再答: b1= (1/4)(1-1/2) b2=(1/4)(1/2-1/3) .. bn=(1/4)(1/n-1/(n+1)) 左和 =右和 Sn=(1/4)(1-1/(n+1))
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