已知数列{an}满足a1=1,an+1=
(1)设bn=
(2)求数列{an}的通项公式an. |
(1)由bn=
,得an=1+24an
,
-1b 2n 24
代入an+1=
(1+4an+1 16
),1+24an
得
=
-1b 2n+1 24
(1+4×1 16
+bn)⇒4
-1b 2n 24
=(bn+3)2,b 2n+1
∴2bn+1=bn+3.…(5分)
∴2(bn+1-3)=bn-3,又b1=
=5,则b1-3=2≠0.…(7分)1+24×1
∴{bn-3}是以2为首项,
为公比的等比数列.…(8分)1 2
(2)由(1)得bn-3=
,∴bn=1 2n-2
+3,…(10分)1 2n-2
则an=
=
-1b 2n 24
×2 3
+1 4n
+1 2n+2
.…(13分)1 3