Question

设Sn是正项数列an的前n项和,对任意n,都有Sn=1/4(an+1)^2,设bn=an/2^n,Tn为数列bn

Answer 1

Tn=1×1/2+3×1/2^2+..+(2n-1)×1/2^n
1/2Tn=1×1/2^2+3×1/2^3+..+(2n-3)×1/2^n+(2n-1)×1/2^(n+1)
两式相减的：1/2Tn=1/2+2×(1/2^2+1/2^3+..+1/2^n)-(2n-1)/2^(n+1)

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

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

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