已知数列{an}的首项,a1=0.5,2an=4a(n-1)+1(n>=2),求它的通项公式

an=2a(n-1)+1/2
an+1/2=2((n-1)+1/2)
数列{an+1/2}是等比数列
首项=a1+1/2=1 q=2
an+1/2=2^(n-1)
an=2^(n-1)-1/2首项=a1+1/2=1没看懂~数列{an+1/2}是等比数列它的首项为a1+1/2a1=0.5所以a1+1/2=1