已知等比數列的通項公式a(n)=3·(1/2)^(n-1),且b(n)=a(3n-2)+a(3n-1)+a(3n) (n∈N*),求證:【b(n)】成等比數列.

熱心網友

b(n)=a(3n-2)+a(3n-1)+a(3n) =[3·(1/2)^(3n-3)]+[3·(1/2)^(3n-2)]+[3·(1/2)^(3n-1)] =[3·(1/2)^(3n-3)][1+1/2+(1/2)^2] =[3·(1/2)^(3n-3)]*(7/4) =3*(7/4)*[(1/2)^3]^(n-1) =(21/4)*(1/8)^(n-1)則b(n+1)=(21/4)*(1/8)^nb(n+1)/b(n)=[(21/4)*(1/8)^n]/[(21/4)*(1/8)^(n-1)]=1/8故【b(n)】成等比數列，公比為1/8