已知数列{an}满足a1=½,a1+a2+.+an=n²an,求其通项an

已知数列{an}满足a1=½,a1+a2+.+an=n²an,求其通项an
已知数列{an}满足a1=½,a1+a2+.+an=n²an,求其通项an

已知数列{an}满足a1=½,a1+a2+.+an=n²an,求其通项an
a1+a2+.+an=n²an
a1+a2+.+a(n-1)=(n-1)²a(n-1)
相减,得
an=n²an-(n-1)²a(n-1)
(n²-1)an=(n-1)²a(n-1)
(n+1)(n-1)an=(n-1)²a(n-1)
(n+1)an=(n-1)a(n-1)
an=(n-1)/(n+1) a(n-1)
=(n-1)/(n+1)*(n-2)/n a(n-2)
=(n-1)/(n+1)*(n-2)/n*(n-3)/(n-1)a(n-3)
=.
=[(n-1)(n-2).2/(n+1)n(n-1).4] a2
=(n-1)(n-2).2×1/(n+1)n(n-1).×4×3 a1
=2/(n+1)n a1
=2/(n+1)n ×1/2
=1/(n+1)n
即
an=1/n(n+1)