解:(1)∵2S
n+1+a
n+1+4S
n+1S
n=0
∴2S
n+1+S
n+1-S
n+4S
n+1S
n=0
即3S
n+1-S
n+4S
nS
n+1=0
两边同时除以S
nS
n+1可得,
![](http://thumb.1010pic.com/pic5/latex/52344.png)
从而可得,
![](http://thumb.1010pic.com/pic5/latex/52345.png)
,
![](http://thumb.1010pic.com/pic5/latex/52346.png)
∴
![](http://thumb.1010pic.com/pic5/latex/52347.png)
以3为首项,以3为公比的等比数列
由等比数列的通项公式可得,
![](http://thumb.1010pic.com/pic5/latex/52348.png)
=3
n∴
![](http://thumb.1010pic.com/pic5/latex/52349.png)
当n≥2时,
![](http://thumb.1010pic.com/pic5/latex/52350.png)
a
1=1不适合上式
故
![](http://thumb.1010pic.com/pic5/latex/52351.png)
(2)由(1)知,
![](http://thumb.1010pic.com/pic5/latex/52343.png)
=(2n+1)•3
n∴T
n=3•3
1+5•3
2+…+(2n-1)•3
n-1+(2n+1)•3
n∴3T
n=3•3
2+5•3
3+…+(2n-1)•3
n+(2n+1)•3
n+1两式相减可得,-2T
n=9+2(3
2+3
3+…+3
n)-(2n+1)•3
n+1整理可得,T
n=n•3
n+1分析:(1)由2S
n+1+a
n+1+4S
n+1S
n=0,可得2S
n+1+S
n+1-S
n+4S
n+1S
n=0即3S
n+1-S
n+4S
nS
n+1=0变形可得,
![](http://thumb.1010pic.com/pic5/latex/52344.png)
,从而可得
![](http://thumb.1010pic.com/pic5/latex/52347.png)
为等比数列,可求S
n,利用
![](http://thumb.1010pic.com/pic5/latex/11310.png)
可求a
n(2)由(1)知,
![](http://thumb.1010pic.com/pic5/latex/52343.png)
=(2n+1)•3
n,利用乘公比错位相减法求和
点评:本题主要考查了利用数列的递推公式求解数列的通项公式,解决问题的关键是根据已知条件构造等比数列,二乘公比错位相减求数列的和是数列部分的重要方法,要注意掌握.