Question

若q>1,bn=2n+1/an,求bn前n项和Tn,an=2的n-1次方,a1=1,q=2

Answer 1

Tn=3+5/2+7/2^2+……+（2n+1)/2^(n-1),①

(1/2)Tn= 3/2+5/2^2+……+(2n-1)/2^(n-1)+(2n+1)/2^n,②

①-②,(1/2)Tn=3+2[1/2+1/2^2+……+1/2^(n-1)]-(2n+1)/2^n

=3+2[1-1/2^(n-1)]-(2n+1)/2^n

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

=5-(2n+5)/2^n,

∴Tn=10-(2n+5)/2^(n-1).