已知数列｛an｝的前n项和Sn满足Sn=(1/4)*(an+1)^2.求：（1）求数列｛an｝的通项公式；（2）设bn=1/(an*an+1),记数列｛bn｝的前n项和为Tn,求Tn

Question

已知数列｛an｝的前n项和Sn满足Sn=(1/4)*(an+1)^2.求：（1）求数列｛an｝的通项公式；（2）设bn=1/(an*an+1),记数列｛bn｝的前n项和为Tn,求Tn

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