Question

设正项数列{an}前项和Sn满足Sn=1/4(an+1)^2 求an

Answer 1

4Sn=(an+1)^2，
n=1时，4S1=(a1+1)^2，4a1=(a1+1)^2，a1=1.
4Sn=(an+1)^2, 4S(n-1)=[a(n-1)+1]^2,
又an=Sn-S(n-1)
上面两式相减得：
4an=(an+1)^2-[a(n-1)+1]^2
(an+1)^2-4an=[a(n-1)+1]^2
(an-1)^2=[a(n-1)+1]^2，(an-1)^2-[a(n-1)+1]^2=0，
（an+a(n-1)）(an-a(n-1)-2)=0,
∵an>0
∴an=a(n-1)+2
所以an是等差数列,首项为1公差为2.
an=1+2(n-1)=2n-1.

Answer 2

S1=a1=1/4(1+a1)^2 ==> a1=1
S2=a1+a2=1/4(1+a2)^2 ==> a2=2 (an>0)
Sn=an+Sn-1=an+1/4[a(n-1)+1]^2 =1/4(an+1)^2 ==>an=1/2a(n-1)+3/2
==>an=3-2*(1/2)^(n-1)

Answer 3

是An+1，还是第n+1项