Question

设正数数列an前n项和为Sn,Sn=二分之一(an+an分之一)，求an的通项公式

Answer 1

a(n)>0.
s(n)
=
[a(n)]^2
+
a(n)/2,
a(1)
=s(1)
=
[a(1)]^2
+
a(1)/2,
0
=
[a(1)]^2
-
a(1)/2
=
a(1)[a(1)-1/2],
a(1)=1/2.
s(n+1)=[a(n+1)]^2
+
a(n+1)/2,
a(n+1)=s(n+1)-s(n)
=
[a(n+1)]^2
+
a(n+1)/2
-
[a(n)]^2
-
a(n)/2,
0
=
[a(n+1)]^2
-
[a(n)]^2
-
a(n+1)/2
-
a(n)/2
=
[a(n+1)+a(n)][a(n+1)-a(n)-1/2],
0
=
a(n+1)-
a(n)-1/2,
a(n+1)
=
a(n)+1/2,
{a(n)}是首项为a(1)=1/2，公差为1/2的等差数列。
a(n)
=
1/2
+
(n-1)/2
=
n/2.

Answer 2

Sn = (1/2)(an + 1/an) (1)
n=1
2(a1)^2 = (a1)^2+a1
a1(a1-1)=0
a1 =1

S(n-1) = (1/2)[a(n-1) +1/a(n-1) ] (2)
(1)-(2)
an = (1/2)(an + 1/an)- (1/2)[a(n-1) +1/a(n-1) ]
1/an - an - a(n-1) -1/a(n-1) =0
a(n-1) - 2an.a(n-1) - an =0
1/an -1/a(n-1) =2
1/an -1/a1 = 2(n-1)
1/an = 2n-1
an = 1/(2n-1)