Question

数列 (9 10:52:38)社数列an是正数组成的数列,其前n项和为Sn,且对所有自然数n,an与2的等差中项等于Sn

Answer 1

(an+2)/2=√2Sn

(an+2)^2=8Sn

an^2+4an+4=8Sn

an-1^2+4an-1+4=8Sn-1

两式相减

(an+an-1)(an-an-1)+4(an-an-1)=8an

[an+a(n-1)][an-a(n-1)]-4[an+a(n-1)]=0

[an+a(n-1)]{[an-a(n-1)]-4}=0

an为正数列；[an+a(n-1)]≠0

an-a(n-1)=4

在(an+2)^2=8Sn,令n=1,a1=2

所以an=4n-2