Question

通项公式an=2^n,bn=log2(an),cn=bn/an,设cn前n项和为Tn,求Tn.

Answer 1

n=n

cn=n/2^n

Tn=1/2+2/2^2+3/2^3+...+(n-1)/2^(n-1)+n/2^n

2Tn=1+2/2+3/2^2+4/2^3+.+n/2^(n-1)

2Tn-Tn=1+1/2+1/2^2+1/2^3+...+1/2^(n-1)-n/2^n

Tn=2-2^(1-n)-n/2^n

=2-(n+2)/2^n