Question

设各项均为正数的数列{an}的前n项和为Sn,已知2sn＝an＋an²

设各项均为正数的数列{an}的前n项和为Sn，已知2Sn＝an+an².
Ⅰ 求{an}的通项公式；
Ⅱ 记bn=a1+a2+…+an分之1，求数列{bn}的前n项和Tn.
最好能加我QQ7937.67474

Answer 1

I.2Sn＝an+an²，①
n=1时2a1=a1+a1^,a1^-a1=0,a1>0,
∴a1=1.
n>1时2S<n-1>=a<n-1>+a<n-1>^,②
①-②，2an=an+an^-a<n-1>-a<n-1>^,
an+a<n-1>=(an+a<n-1>)(an-a<n-1>),an+a<n-1>>0,
∴an-a<n-1>=1,
∴{an}是等差数列，an=n.
II.a1+a2+……+an=n(1+n)/2,
∴bn=2/[n(1+n)]=2[1/n-1/(1+n)],
∴Tn=2[1-1/2+1/2-1/3+……+1/n-1/(1+n)]
=2[1-1/(1+n)]
=2n/(1+n).