(2011•广安二模)已知数列{an}满足a1=2,an+1=3an+2n-1(n∈N*).
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(2011•广安二模)已知数列{an}满足a1=2,an+1=3an+2n-1(n∈N*).
(1)求证数列{an+n}是等比数列,并求an
(2)若数列{bn}中1>2=6,前n项和为Tn,且9Tn-a=(an+n)bn(n∈N*),求数列{bn}的通项公式.
(1)求证数列{an+n}是等比数列,并求an
(2)若数列{bn}中1>2=6,前n项和为Tn,且9Tn-a=(an+n)bn(n∈N*),求数列{bn}的通项公式.
(1)由题设得an+1+(n+1)=3(an+n)∵a1+1=3
∴{an+n}是首项为3,公比为3的等比数列,
∴an+n=3n
∴an=3n-n
(2)∵9Tn-a=(an+n)bn(n∈N*),即32Tn-2n=3nbn
∴2Tn-2n=nbn ①
由①得2Tn+1-2(n+1)=(n+1)bn+1 ②,
②-①得2(bn+1-1)=(n+1)bn+1-nbn
即(n-1)bn+1-nbn+2=0 ③
由③得nbn+2-(n+1)bn+1+2=0 ④
④-③得nbn+2-2nbn+1+nbn=0
∴bn+2-bn+1=bn+1-bn
∴{bn}是等差数列
由9b1-1=3b1 得b1=2,又∵b2=6
∴公差d=4
∴bn=b1+(n-1)d=4n-2
∴{an+n}是首项为3,公比为3的等比数列,
∴an+n=3n
∴an=3n-n
(2)∵9Tn-a=(an+n)bn(n∈N*),即32Tn-2n=3nbn
∴2Tn-2n=nbn ①
由①得2Tn+1-2(n+1)=(n+1)bn+1 ②,
②-①得2(bn+1-1)=(n+1)bn+1-nbn
即(n-1)bn+1-nbn+2=0 ③
由③得nbn+2-(n+1)bn+1+2=0 ④
④-③得nbn+2-2nbn+1+nbn=0
∴bn+2-bn+1=bn+1-bn
∴{bn}是等差数列
由9b1-1=3b1 得b1=2,又∵b2=6
∴公差d=4
∴bn=b1+(n-1)d=4n-2
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