问题
解答题
已知数列{an},{bn}满足a1=
(1)求数列{an},{bn}的通项公式; (2)求数列{anbn}的前n项和Tn; (3)若数列{cn}满足bn=
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答案
(1)由已知,对任意m,n∈N*,
有am+n=am•an,bm+n=bm+bn.
取m=1,得an+1=a1an=
an,bn+1=b1+bn=-1 2
+bn.1 2
所以数列{an},{bn}分别为等比,等差数列.
∴an=
•(1 2
)n-1=(1 2
)n1 2
bn=-
+(n-1)(-1 2
)=-1 2
…(4分)n 2
(2)Tn=(-
)(1 2
)1+(-1 2
)(2 2
)2+(-1 2
)(3 2
)3+…+(-1 2
)(n 2
)n1 2
Tn=(-1 2
)• (1 2
) 2+(- 1 2
)•(2 2
)3+…+(-1 2
)•(n 2
)n+11 2
两式相减,
Tn=-1 2
-1 22
[(1 2
)2+(1 2
)3+…+(1 2
)n]+1 2
•(n 2
)n+11 2
并化简得Tn=n×(
)n+1+(1 2
)n-1.…(8分)1 2
(3)由bn=
,4cn+n 3cn+n
得cn=-
.…(10分)n2+2n 3n+8
∵cn+1-cn=-
<0.3n2+19n+24 (3n+8)(3n+11)
∴数列{cn}为递减数列,cn的最大值为c1.
故存在M=1,使得对任意n∈N*,cn≤c1恒成立…