Question

已知数列{an}满足a1=1,an=a(n-1)/3a(n-1)+1,(n>=2,n属于N*) 设bn=an×a（n+1）（n属于N*）求数列{bn}的前n

求数列{bn}的前n项和

Answer 1

an=a(n-1)/(3a(n-1)+1)
倒一下：
1/an=(3a(n-1)+1)/a(n-1)=3+1/a(n-1)
所以1/an-1/a(n-1)=3 1/a1=1
所以1/an是等差数列
1/an=1/a1+3*(n-1)=3n-2
an=1/(3n-2)
所以bn=an*a(n+1)=1/(3n-2)*1/(3n+1)=1/3[1/(3n-2)-1/(3n+1)]
所以bn前n项和为
Sn=b1+b2……+bn
=1/3(1/1-1/4+1/4-1/7+1/7-1/10……+1/(3n-2)-1/(3n+1))
=1/3(1-1/(3n+1))
=1/3*3n/(3n+1)
=n/(3n+1)