设{an}是正项数列,它的前n项和Sn满足:4Sn=(an-1)?(an+3),则a1005=________.
网友回答
2011
解析分析:把数列仿写一个,两式相减,合并同类型,用平方差分解因式,约分后得到数列相邻两项之差为定值,得到数列是等差数列,公差为2,取n=1代入4Sn=(an-1)(an+3)得到首项的值,写出通项公式.从而得到a1005.
解答:∵4Sn=(an-1)(an+3),∴4sn-1=(an-1-1)(an-1+3),两式相减得整理得:2an+2an-1=an2-an-12,∵{an}是正项数列,∴an-an-1=2,∵4Sn=(an-1)(an+3),令n=1得a1=3,∴an=2n+1,∴a1005=2×1005+1=2011.故