等差数列{an}的前n项和为Sn,满足a3=7,且a5+a7=26,求：令bn=1∕(an²-4﹚,求数列bn

设公差为d,
a5+a7
=a3+2d+a3+4d
=2a3+6d
=26
代入a3=7
d=2
a1=a3-2d=3
an=2n+1
bn=1/(an²-4)
=1/［(an+2)(an-2)］
=1/4×［1/(an-2)-1/(an+2)］
b1=1/4×（1-1/5）
Sbn=1/4×{(1-1/5)+(1/3-1/7)+(1/5-1/9)+（1/7-1/11）+...［1/(2n-1)-1/(2n+3)］}
=1/4×［1+1/3-1/(2n+1)-1/(2n+3)］
=1/4×{4/3-（4n+4)/［(2n+1)(2n+3)］}
=1/3-（n+1)/(4n²+8n+3)