题目内容
已知等差数列{an}的前n项和为Sn,n∈N*,且a2=3,点(10,S10)在直线y=10x上.
(1)求数列{an}的通项公式;
(2)设bn=2an+2n,求数列{bn}的前n项和Tn.
(1)求数列{an}的通项公式;
(2)设bn=2an+2n,求数列{bn}的前n项和Tn.
(1)an=2n-1(2)
×4n+n2+n-


(1)设等差数列{an}的公差为d,
∵点(10,S10)在直线y=10x上,∴S10=100,
又∵a2=3,∴
解得
∴an=2n-1.
(2)∵bn=2an+2n=
×4n+2n,
∴Tn=b1+b2+…+bn=
(4+42+…+4n)+2(1+2+…+n)=
+n2+n
=
×4n+n2+n-
.
∵点(10,S10)在直线y=10x上,∴S10=100,
又∵a2=3,∴


(2)∵bn=2an+2n=

∴Tn=b1+b2+…+bn=


=



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