一道很简单的求通项公式题

设bn+x=k(b(n-1)+x)
则bn=b(n-1)*k+(k-1)x
所以(k-1)x=2
若k≠1
x=2/(k-1)
所以bn+2/(k-1)是公比为k的等比数列
bn+2/(k-1)=[b1+2/(k-1)]*k^(n-1)
=[(k+1)/(k-1)]*k^(n-1)
bn=[(k+1)/(k-1)]*k^(n-1) - 2/(k-1)
=[(k+1)*k^(n-1)-2]/(k-1）
若K=1
bn=bn-1+2,是等差数列
则bn=2n-1