已知数列{an}中,a1=-1,an+a(n+1)+4n+2=0若bn=an+2n,求证{bn}为等比数列求{an}的通项公式an
问题描述：
已知数列{an}中,a1=-1,an+a(n+1)+4n+2=0
若bn=an+2n,求证{bn}为等比数列
求{an}的通项公式an
答
bn+1= an+1 +2n+2
b1=a1+2=1
an+an+1 +4n+2 =0
bn+bn+1 =0
bn+1 =-bn
{bn}为等比数列 公比为 -1
bn=（-1）^(n-1)
an+2n=bn=（-1）^(n-1)
an= （-1）^(n-1) -2n