Question

已知函数 fx=x/(3x+1)，数列an满足a1=1, a n+1=f(an),n属于N+。(1

已知函数 fx=x/(3x+1)，数列an满足a1=1, a n+1=f(an),n属于N+。(1)求证数列1/an是等差数列(2)记sn=a1a2+a2a3+……+ana n+1，求sn

Answer 1

您好!
1.a[n+1]=f(an)=an/(3an + 1)
两边同时倒数,1/a[n+1]=(3an+1)/an=3 + 1/an即1/a[n+1] - 1/an =3
故{1/an}是等差数列,1/an=1/a1+(n-1)*3=3n-2, an=1/(3n-2)
2.bn=an*an+1=1/(3n+1)(3n-2)= 1/3[1/(3n-2) - 1/(3n+1)]
Sn=b1+b2+……+bn
=1/3(1-1/4)+1/3(1/4-1/7)+……+1/3[1/(3n-2) - 1/(3n+1)]
=1/3(1-1/(3n+1))
=n/(3n+1)
如有疑问,请在线追问,随时为您解答!

追问

打错了，是a n+1,,n+1是a的n+1项

不好意思，能再解吗

亲还在吗