令bn=(2^(n-1)+1)/((3n-2)an)数列{bn^2}的前n项和为tn,证明对于任意的n∈N+,都有tn

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(1) = 2/4 = 1/2,
t(1) = 1/4 = 3/12 < 5/12.
n>=2时,b(n)= [2^(n-1)+1]/[(3n-2)a(n)] = 1/(3n-2).
[b(n)]^2 = 1/(3n-2)^2,
t(2) = [b(1)]^2 + [b(2)]^2 = 1/4 + 1/16 = 5/16 < 5/12.
n>=3时,3(n-1)-2 >= 4.
[b(n)]^2 = 1/(3n-2)^2 < 1/[(3n-2)(3n-5)] =(1/3) [1/(3n-5) - 1/(3n-2)]
t(n) = [b(1)]^2 + [b(2)]^2 + [b(3)]^2 + ...+ [b(n-1)]^2 + [b(n)]^2
= 1/4 + 1/16 + (1/3)[1/4 - 1/7 + 1/7 - 1/10 + ...+ 1/(3n-8)-1/(3n-5) + 1/(3n-5) - 1/(3n-2)]
= 5/16 + (1/3)[1/4 - 1/(3n-2)]
< 5/16 + 1/12
= 15/48 + 4/48
= 19/48
< 20/48
= 5/12.
综合,有,t(n) < 5/12

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