n=1/(an·a(n+1)=[1/an-1/a(n+1)]/4
Tn=b1+b2+...+b(n-1)
=(1/4)[1/a1-1/a2+1/a2-1/a3+...+1/a(n-1)-1/an]
=(1/4)[1/a1-1/an]
=(n-1)/(4n-3)
数列an=4n-3,bn=1/(an·a(n+1),Tn为数列{bn}前n-1项和,求Tn.
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