问题
解答题
设等比数列{an}的前n项和为Sn,已知an+1=2Sn +2(n∈N*). (I)求数列{an}的通项公式; (Ⅱ)求数列{
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答案
(I)由an+1=2Sn +2(n∈N*)可得an=2sn-1+2(n≥2)
两式相减可得,an+1-an=2an
即an+1=3an(n≥2)
又∵a2=2a1+2,且数列{an}为等比数列
∴a2=3a1
则2a1+2=3a1
∴a1=2
∴an=2•3n-1
(II)由(I)知,an=2•3n-1
∴
=n+1 2an n+1 4•3n-1
Tn=
+2 4•30
+3 4•31
+…+4 4•32 n+1 4•3n-1
Tn=1 3
+2 4•31
+…+3 4•32
+n 4•3n-1 n+1 4•3n
两式相减可得,
Tn=2 3
+2 4•30
+1 4•3
+…+1 4•32
-1 4•3n-1 n+1 4•3n
=
+1 2
×1 4
-
(1-1 3
)1 3n-1 1- 1 3
=n+1 4•3n
-5 8 2n+5 8•3n