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

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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

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
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

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
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).

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