您好!
1.an+1=f(an)=an/(3an + 1)
两边同时倒数,1/an+1=(3an+1)/an=3 + 1/an即1/an+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)
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