已知数列{an}是等差数列,且a1=2,
(1)求数列{an}的通项公式; (2)令bn=an•3n+n,求数列{bn}的前n项和Tn. |
(1)∵a1=2,
an+1-1 2
an=2(cos21 2
-sin2π 6
)=2cosπ 6
π=11 3
∴an+1-an=2
∴数列{an}是以2为首项,以2为公差的等差 数列
∴an=2+2(n-1)=2n
(2)∵bn=an•3n+n=2n•3n+n
∴Tn=2(1•3+2•32+…+n•3n)+(1+2+…+n)
∴3Tn=2( 1•32+2•33+…+n•3n+1)+3(1+2+…+n)
两式相减可得,-2Tn=2(3+32+33+…+3n-n•3n+1)-2•n(1+n) 2
=2•
-n(n+1)3(1-3n) 1-3
=3n+1-3-n(n+1)
∴Tn=n(n+1)+3-3n+1 2