一已知数列《2^(n-1)*an》前N项和Sn=9-6n,(1)求an(2)设Bn=n*(3-㏒2(∣an∣/3))求Bn?
(1)
n=1时,an=S1=3
n>1时
2^(n-1)*an=Sn-S(n-1)=-6
so: an=-6/2^(n-1)
(2)n=1时
B1=3-log2(1)=3
n>1时漏银友,搏芦an<返槐0,|an|=-an=6/2^(n-1)
Bn=n*(3-log2(2/2^(n-1)))
=n*(3-log2(2^(2-n)))
=n*(3-(2-n))
=n*(n+1)