拿第一題舉個(gè)例。等式兩邊+k，則
a(n+1)+k=3an+2+k=3(an+(k+2)/3)
a(n+1)+k???/???an+(k+2)/3??=3
令k=(k+2)/3，解得k=1
所以a(n+1)+1???/???an+1=3
所以｛an+1｝是公比為3的等比數(shù)列
an+1=(a1+1)*3^n-1，整理可得an通項(xiàng)公式

推廣到一般，給出a1以及a(n+1)=xan+y(x,y為常數(shù))即可求an通項(xiàng)公式
兩邊+k得
a(n+1)+k=xan+y+k=x(an+(y+k)/x)
a(n+1)+k???/???an+(y+k)/x?=x
令k=y+k/x，解得k=y/x-1
所以｛an+y/(x-1)｝是公比為x的等比數(shù)列
an+y/(x-1)=(a1+y/(x-1))*x^n-1
an=(a1+y/(x-1))*x^(n-1)-y/(x-1)

詳見(jiàn)：

https://www.bilibili.com/video/BV11Z4y1w7Px