已知f(x)=2x/(x+2),a1=1,a(n+1)=f(an),bn=ana(n+1)
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热心网友
时间：2024-10-21 09:42
a(n+1)=f(an)=2an/(an+2)
a(n+1)(an+2)=2an
a(n+1)an+2a(n+1)=2an 两边除a(n+1)an得
1+2/an=2/a(n+1)
1/a(n+1)-1/an=1/2
所以{1/an)是等差数列
2)
因为{1/an)公差是1/2
则1/an=1/a1+(n-1)d
=1/1+(n-1)*1/2
=1+n/2-1/2
=(n+1)/2
an=2/(n+1)
3)
bn=ana(n+1)
=2/(n+1)*2/(n+2)
=4*1/(n+1)*(n+2)
=4*(1/(n+1)-1/(n+2))
所以Tn=b1+b2+b3.....+bn
=4(1/2-1/3+1/3-1/4+...........+1/(n+1)-1/(n+2))
=4(1/2-1/(n+2))
=4(n+2-2)/2(n+2)
=2n/(n+2)追问为什么ana(n+1)=2(an-a(n+1))？
追答2(an-a(n+1))
4*(1/(n+1)-1/(n+2))
=4*(n+2-(n+1)/(n+1)(n+2)
=4(n+2-n-1)/(n+1)(n+2)
=4/(n+1)(n+2)
=ana(n+1)
