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
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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](/uploads/image/z/2427541-61-1.jpg?t=%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E6%BB%A1%E8%B6%B3a3%3D7%2C%E4%B8%94a5%2Ba7%3D26%2C%E6%B1%82%EF%BC%9A%E4%BB%A4bn%3D1%E2%88%95%28an%26%23178%3B-4%EF%B9%9A%2C%E6%B1%82%E6%95%B0%E5%88%97bn%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C)
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
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
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)
a3=7,a5+a7=26,因为是等差,所以2a6=26,a6=13,d=2,a1=3 ,an=2n+1
bn=1/4n^2+4n-3 =1/4(1/2n-1-1/2n+3)
Sbn=1/4(1-1/5+1/3-1/7+1/5-1/9+...-1/2n-3-1/2n+3)=1/3-n/2n-3
a5+a7=2a6=26
a6=13,a6-a3=3d=6
d=2,an=1+2n
bn=1/[(2n-1)(2n+3)]=(1/4)x[(1/2n-1)-(1/2n+3)]
Sbn=(1/4)x[4/3-1/(2n+1)-1/(2n+3)]
=1/3-(n+1)/[(2n+1)(2n+3)]
因为a5+a7=26 所以2a6=26 a6=13
a3+3d=a6 所以d=2 an=2n+1
所以bn=1/(4乘n平方+4n-3)=1/(2n-1)(2n+3)
bn前n项和=1/(1*5)+1/(3*7)+1/(7*11)+·······+1/(2n-1)2n+3)
=1/4(1-1\5+1/5-1/7+1/7-1/11+····...
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因为a5+a7=26 所以2a6=26 a6=13
a3+3d=a6 所以d=2 an=2n+1
所以bn=1/(4乘n平方+4n-3)=1/(2n-1)(2n+3)
bn前n项和=1/(1*5)+1/(3*7)+1/(7*11)+·······+1/(2n-1)2n+3)
=1/4(1-1\5+1/5-1/7+1/7-1/11+······+2n-1-2n-3)
=1/4(1-2n-3)
=-1/2-n/2
我确定这样做 过程简写了
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因为a5+a7=26 所以2a6=26 a6=13
a3+3d=a6 所以d=2 an=2n+1
所以bn=1/(4乘n平方+4n-3)=1/(2n-1)(2n+3)=1/(2n-1)(2n+3)=(1/4)/[1/(2n-1)-1/(2n+3)]
bn前n项和=(1/4)[1+1/3-1/(2n+1)-1/(2n+3]=1/3+(n+1)/(2n+1)(2n+3)