问题
解答题
已知数列{an}的前n项和Sn=-an-(
(1)令bn=2nan,求证数列{bn}是等差数列,并求数列{an}的通项公式; (2)令cn=
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答案
(1)在Sn=-an-(
)n-1+2中令n=1可得s1=-a1-1+2=a1即a1=1 2 1 2
当n≥2时an=Sn-Sn-1=-an+an-1+(
)n-11 2
∴2an=an-1+(
)n-1即2nan=2n-1an-1+11 2
∵bn=2nan,
∴bn-bn-1=1即当n≥2时bn-bn-1=1
又∵b1=2a1=1
∴数列{bn}是首项和公差均为1的等差数列.
∴bn=1+(n-1)×1=n=2nan
∴an=n 2n
(2)由(1)得cn=(n+1)(
)n,1 2
∴Tn=2×
+3×(1 2
)2+4×(1 2
)3+…+(n+1)(1 2
)n ①1 2
Tn=2×(1 2
)2+3×(1 2
)3+4×(1 2
)4+…+(n+1)(1 2
)n+1 ②1 2
由①-②得
Tn=1+(1 2
)2+(1 2
)3+…+(1 2
)n-(n+1)(1 2
)n+1=1 2
-3 2 n+3 2n
∴Tn=3-n+3 2n