设数列{an}为等比数列,数列{bn}满足bN=na1+(n-1)a2+…+2an-1+an,n属于正整数.已知b1=m,b2=3m/2,其中m

设数列{an}为等比数列,数列{bn}满足bN=na1+(n-1)a2+…+2an-1+an,n属于正整数.已知b1=m,b2=3m/2,其中m
设数列{an}为等比数列,数列{bn}满足bN=na1+(n-1)a2+…+2an-1+an,n属于正整数.已知b1=m,b2=3m/2,其中m

设数列{an}为等比数列,数列{bn}满足bN=na1+(n-1)a2+…+2an-1+an,n属于正整数.已知b1=m,b2=3m/2,其中m
m不等于0,求数列an的首项和
令等比数列an=a1q^(n-1),
b1=a1=m,
b2=2a1+a2=(3/2)m,
联立争得a2=-(1/2)m,
则q=a2/a1=-1/2,
则an是一个以m为首项,-1/2为公比的等比数列!