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*)

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