Question

已知正数数列{an}的前n项和为Sn,且有Sn=1/4×（an+1）²,数列B1,b2-b1,b3-b2,..

Answer 1

Sn=(an+1)²/4 ∴4Sn=(an+1)²………………①

n=1时, 4a1=(a1+1)², 解得a1=1

n>1时, 4S(n-1)=(a(n-1)+1)²………………②

①-②得 4an=(an+1)²-(a(n-1)+1)²

∴(an+1)²-4an-(a(n-1)+1)²=(an²-2an+1)-(a(n-1)+1)²=(an-1)²-(a(n-1)+1)²=0

∴(an-1)²-(a(n-1)+1)²=[an-1+a(n-1)+1][an-1-a(n-1)-1]=[an+a(n-1)][an-a(n-1)-2]=0

∵{an}为正, ∴an=a(n-1)+2

∴{an}是首项为1,公差为2的等差数列

∴an=1+2(n-2)=2n-1