Question

第五题求解，解答过程请给出，谢谢了

Answer 1

1）2√S1=a1+1
a1²-2a1+1=0
a1=1
2）Sn=(an+1)²/4
S(n-1)=[a(n-1)+1]²/4
an=Sn-S(n-1)=(an+1)²/4-[a(n-1)+1]²/4
[an+a(n-1)][an-a(n-1)-2]=0
∵an+a(n-1)>0
∴an-a(n-1)-2=0
an-a(n-1)=2
因此，{an}是公差为2的等差数列
an=a1+(n-1)*2=2n-1
3）bn=1/[an*a(n-1)]=1/[(2n-1)(2n+1)]=1/2[1/(2n-1)-1/(2n+1)]
Tn=b1+b2+b3+...+bn
=1/2*[1-1/3+1/3-1/5+1/5-1/7+...+1/(2n-1)-1/(2n+1)]
=1/2*[1-1/(2n+1)]
=n/(2n+1)