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
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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的前n项和
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)
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)
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