已知等差数列an绝对值,an+1=2an-3n+1.则第五项a5=

已知等差数列an绝对值,an+1=2an-3n+1.则第五项a5=
已知等差数列an绝对值,an+1=2an-3n+1.则第五项a5=

已知等差数列an绝对值,an+1=2an-3n+1.则第五项a5=
an=3n,a5=15

a(n+1)=2an-3n+1
an=2a(n-1)-3(n-1)+1
相减：
a(n+1)-an=2an-3n+1-2a(n-1)+3(n-1)-1=2an-2a(n-1)+3
a(n+1)-an=2[an-a(n-1)]+3
a(n+1)-an+3=2[an-a(n-1)+3]
an-a(n-1)+3=(a2-a1+3)2^(n-2)
表达清楚1点，看不懂呀？