已知数列{an}满足a1=4,an+1an+6an+1-4an-8=0,记bn=6an−2.
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已知数列{an}满足a1=4,an+1an+6an+1-4an-8=0,记bn=
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a
(1)∵bn=
6 an−2, ∴an= 6 bn+2, 又∵an+1an+6an+1-4an-8=0, ∴( 6 bn+1+2)( 6 bn+2)+6( 6 bn+1+2)-4( 6 bn+2)-8=0, 整理得bn+1=4bn+3 bn+1+1=4(bn+1) ∴{bn+1}是首项是b1+1= 6 4−2+1=4,公比为4的等比数列, ∴bn+1=4×4n-1=4n, ∴bn=4n-1. (2)anbn=( 6 bn+2)bn=2bn+6=2×4n+4=22n+1+4, ∴sn=(23+25+…+22n+1)+4n= 23(1−4n) 1−4+4n= 22n+3+12n−8 3.
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