Question

各项为正数的数列{an}的前n项和为Sn，且满足：Sn=（1/4）(an)^2+(1/2)an+1/4

已求得an=2n-1
设函数f（n）=an，n为奇数时;f（n）=f（n/2）(n为偶数)，cn=f（2^n+4）,求数列{cn}的前n项和Tn

Answer 1

a(n)=2n-1
f(2n-1)=a(2n-1)=2(2n-1)-1=4n-3
f[2^k(2n-1)]=f(2n-1)=4n-3
c(1)=f(2+4)=f(3)=f(2*2-1)=4*2-3=5,
c(2)=f(4+4)=f(1)=f(2*1-1)=4*1-3=1,
n>2时,
c(n)=f[2^n+4]=f[2^(n-2)+1]=f{2*[2^(n-3)+1]-1}=4[2^(n-3)+1]-3=1+2^(n-1)
t(1)=c(1)=5
t(2)=c(1)+c(2)=6
n>2时,
t(n)=c(1)+c(2)+[1+2^(3-1)] + [1+2^(4-1)] + ... +[1+2^(n-1)]
=1+[1+2^(1-1)]+[1+2^(2-1)]+[1+2^(3-1)]+...+[1+2^(n-1)]
=1+n+[1+2+...+2^(n-1)]
=1+n+[2^n-1]
=n+2^n,n=3,4,...