Question

请在这里概述您的问题各项为正数的数列{an}的前n项和为Sn，且满足Sn=1/4an^2+1/2an+1/4(n∈N*)

（1）求an
（2）设函数 {an，n为奇数，
f(n)= { Cn=f(2^n+4)(n∈N*)，求数列{Cn}的前n项和Tn
{f(n/2)，n为偶数，

Answer 1

(1) Sn=0.25an²+0.5an+0.25=0.25(an+1)²
即S(n-1)+an=0.25(an+1)² => S(n-1)=0.25(an-1)²
而S(n-1)=0.25(a(n-1)+1)² => (an-1)²=(a(n-1)+1)²
∵a1=s1=0.25(a1+1)² => a1=1 => (a2-1)²=4 =>a2=3
由此结合正项数列知an≥1,∴an-1=a(n-1)+1 => an=a(n-1)+2
=> an=2n-1
(2) ∴Cn=f(4(2^(n-2)+1))=f(2(2^(n-2)+1))=f(2^(n-2)+1)
=a[(2^(n-2)+1)]=2^(n-1)+1，n≥2
而C1=f(2+4)=f(3)=a3=5
∴Tn=C1+C2+....+Cn=5+2+1+2²+1+...+2^(n-1)+1
=2^n+n+2