已知数列{an}的前n项和是Sn=32*
(1)
An=Sn-S(n-1)
=32n-n^2-32n+32+n^2-2n+1
=-2n+33
设Bn=A(2n)-1=-2(2n)+33-1=-4n+32
则Bn+1=A(2n+1)-1=-2[2(n-1)]+33-1=-2(2n-2)+32=-4n+36
所以{a(2n)-1}是公差为4的等差数列
(2)
由(1)可得
-2n+33>0 n<33/2,n≤16 。
所以
n≤16时,
|An|=|-2n+33 |=-2n+33
Tn =|32n-n^2|=-n^2+32n
当n>16时,
|An|=|-2n+33 |=2n-33
Tn =|Sn-S16|+|S16|
=|32n...全部
(1)
An=Sn-S(n-1)
=32n-n^2-32n+32+n^2-2n+1
=-2n+33
设Bn=A(2n)-1=-2(2n)+33-1=-4n+32
则Bn+1=A(2n+1)-1=-2[2(n-1)]+33-1=-2(2n-2)+32=-4n+36
所以{a(2n)-1}是公差为4的等差数列
(2)
由(1)可得
-2n+33>0 n<33/2,n≤16 。
所以
n≤16时,
|An|=|-2n+33 |=-2n+33
Tn =|32n-n^2|=-n^2+32n
当n>16时,
|An|=|-2n+33 |=2n-33
Tn =|Sn-S16|+|S16|
=|32n-n^2-32*16+16^2|+32*16-16^2
=|32n-n^2-16^2|+16^2
=|-(n-16)^2|+16^2
=(n-16)^2+16^2
=n^2-32n+512
当n≤16时,Tn =-n^2+32n
当n>16时,Tn =n^2-32n+512 。
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