已知等比数列{an}的通项公式an=3*(1/2)^(n-1)且:bn=a(3n-2)+a(3n-1)+a(3n),求证：数列{bn}成等比数列
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bn=a(3n-2)+a(3n-1)+a(3n)=3*(1/2)^(3n-2-1)+3*(1/2)^(3n-1-1)+3*(1/2)^(3n-1)=3*(1/2)^(3n-3)+3*(1/2)^(3n-2)+3*(1/2)^(3n-1)=3*(1/2)^(3n-1)[(1/2)^-2+(1/2)^-1+1]=3*(1/2)^(3n-1)*(4+2+1)=21*(1/2)^(3n-1)
b(n+1)=21*(1/2)^[3(n+1)-1]=21*(1/2)^(3n+2)
b(n+1)/bn=[21*(1/2)^(3n+2)]/21*(1/2)^(3n-1)]=(1/2)^3=1/8 为常数
所以bn是等比数列