已知数列{an}的前n项和Sn>1,且6Sn=(an+1)(an+2),n属于N*求(1)通项an(2)Sn急!!!!!!!!!!!!!!

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已知数列{an}的前n项和Sn>1,且6Sn=(an+1)(an+2),n属于N*求(1)通项an(2)Sn急!!!!!!!!!!!!!!已知数列{an}的前n项和Sn>1,且6Sn=(an+1)(a

已知数列{an}的前n项和Sn>1,且6Sn=(an+1)(an+2),n属于N*求(1)通项an(2)Sn急!!!!!!!!!!!!!!
已知数列{an}的前n项和Sn>1,且6Sn=(an+1)(an+2),n属于N*求(1)通项an(2)Sn
急!!!!!!!!!!!!!!

已知数列{an}的前n项和Sn>1,且6Sn=(an+1)(an+2),n属于N*求(1)通项an(2)Sn急!!!!!!!!!!!!!!
(I)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.
Sn=3(n+1)n/2-n