数列{an}=2n+1,使bn=1/an^2-1,求bn前n项和,

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$数列{an}=2n+1,使bn=1/an^2-1,求bn前n项和,$

n=1/(an^2-1)
=1/(2n+1-1)(2n+1+1)
=1/2n(2n+2)
=1/4n(n+1)
=1/4[1/n-1/(n+1)]
∴bn前n项和=1/4(1-1/2+1/2-1/3+1/3-1/4+..+1/n-1/(n+1)]
=1/4[(1-1/(n+1)]
=n/4(n+1)
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