Question

各项均为正数的数列{an}的前n项和为Sn满足2Sn=an(an+1),n∈N*,求an

Answer 1

2Sn=(an+1)*an=an^2+an
2S(n+1)=a(n+1)^2+a(n+1)
2S(n+1)-2Sn=2a(n+1)=[a(n+1)^2+a(n+1)]-(an^2+an)
整理得：a(n+1)^2-a(n+1)-an^2-an=0
即[a(n+1)^2-an^2]=[a(n+1)+an],[a(n+1)-an]*[a(n+1)+an]=a(n+1)+an
因为任意an>0,等式两边a(n+1)+an可消,于是有a(n+1)-an=1
2Sn=an^2+an,则2a1=a1^2+a1,a1=1
an=1+(n-1)*d=1+(n-1)*1=n
可能过程有些麻烦了