Question

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

Answer 1

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)

Answer 2

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)]

Answer 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)=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)

Answer 4

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

Answer 5

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