Question

若等差数列an的前n项和Sn=[(1+an)/2]^2,n属于N*,求an的通项公式

Answer 1

S(n)=[(1+A(n))/2]^2

S(n-1)=(1+A(n-1)/2]^2

A(n)=S(n)-S(n-1)

=(2(A(n)-A(n-1))+(A(n))^2-(A(n-1))^2)/4

n>=2

(An+An-1)(An-An-1)=2(An+An-1)

An-A(n-1)=2

A1=S1=((1+A1)/2)^2

A1=1

An=2n-1