Question

高中数学数列题：已知各项均为正数的数列{an}的前n项和sn满足sn>1,且6sn=(an+1)(an+2),n属于正整

Answer 1

由a1=S1=1/6(a1+1)(a1+2),解得a1=1或a1=2,

由假设a1=S1>1,因此a1=2,

又由a(n+1)=S(n+1)-Sn=1/6(a(n+1)+1)(a(n+1)+2)-1/6(an+1)(an+2),

得(a(n+1)+an)(a(n+1)-an-3)=0,

即a(n+1)-an-3=0或a(n+1)=-an,因an>0,故a(n+1)=-an不成立,舍去.

因此a(n+1)-an=3,从而{an}是公差为3,首项为2的等差数列,

故{an}的通项为an=3n-1.