Question

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

Answer 1

a1+a2+....+an=n²an......(1) (n>=2)
a1+a2+....+a(n-1)=(n-1)²a(n-1)......(2) (n>=2)
(1)-(2)=an=n²an-(n-1)²a(n-1)
(n-1)²a(n-1)=(n-1)(n+1)an (n>=2)
(n-1)a(n-1)=(n+1)an (n>=2)
an/a(n-1)=(n-1)/(n+1)
a2/a1=1/3
a3/a2=2/4
....
an/a1=a2/a1*a3/a2*...*an/a(n-1)=2/n(n+1)
an=1/n(n+1) (n>=2)
a1=½,适应上式
所以an=1/n(n+1)

Answer 2

An=1/n(n+1)
Sn=n/(n+1)=n²An
裂项相加得Sn=1-1/n+1