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]
![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](/uploads/image/z/2872523-11-3.jpg?t=%E5%B7%B2%E7%9F%A5%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%EF%BD%9Ban%EF%BD%9D%E6%BB%A1%E8%B6%B3%EF%BC%9Aa3%3D7%2Ca5%2Ba7%3D26.%EF%BD%9Ban%EF%BD%9D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn.%E6%B1%82%E4%BB%A4bn%3D1%2F%28an%29%5E2-1%2C%E6%B1%82%EF%BD%9Bbn)
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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