an+an-1=3(an-1+an-2）
設(shè)bn=an+an-1那么bn-1=an-1+an-2
得｛bn｝為首項(xiàng)為a2+a1,公比為3等比數(shù)列
得an+an-1=10*(3∧n-2)
設(shè)·an=(3∧n-2)*10*cn,an-1=(3∧n-3)*10*cn-1
3cn+cn-1=3
將cn,cn-1換成·x,解得x=3/4,3cn+cn-1-3/4=3-3/4,轉(zhuǎn)化得3[（cn)-9/4]=-[(cn-1)-9/4]
｛（cn)-9/4｝為首項(xiàng)(c1)-9/4公比為-1/3等比數(shù)列
得cn通項(xiàng)帶入an=(3∧n-2)*10*cn得·an