an=Sn-S(n-1)=(1/2)(1-an)-(1/2)[1-a(n-1)]2an=1-an-1+a(n-1)an/a(n-1)=1/3a1=S1=(1/2)(1-a1)2a1=1-a1a1=1/3an=a1q^(n-1)=(1/3)(1/3)^(n-1)=1/3^nbn=nan=n/3^n令x=b1+b2+b3+……+bn=1/3+2/3^2+3/3^3+……+n/3^n(1/3)...
{an}中 sn=1/2(1-an) 若{bn}满足 bn=nan 求证 b1+b2+b3+······+bn 小于3/
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