The sum of the first n terms of the sequence {an} and {BN} is denoted as an and BN respectively. It is known that an = - n-3 / 2,4bn-12an = 13N (n ∈ natural number) (base n after a, B, a, b) ① Find the analytic expression of B base n about N and the general term formula of the sequence {B base n}. ② let C base n = (1 / 2) ∧ 2B base n-3a base n, prove that {C base n} is an equal ratio sequence, and find the first n term and C base n of the sequence {C base n} (thank you for the detailed process

The sum of the first n terms of the sequence {an} and {BN} is denoted as an and BN respectively. It is known that an = - n-3 / 2,4bn-12an = 13N (n ∈ natural number) (base n after a, B, a, b) ① Find the analytic expression of B base n about N and the general term formula of the sequence {B base n}. ② let C base n = (1 / 2) ∧ 2B base n-3a base n, prove that {C base n} is an equal ratio sequence, and find the first n term and C base n of the sequence {C base n} (thank you for the detailed process

1. An = - n-3 / 2 = - 5 / 2 - (n-1), that is, the arithmetic sequence of the first term A1 = - 5 / 2 and the tolerance d = - 1,
——》An=na1+n(n-1)d/2=-5n/2-n(n-1)/2=-(n^2+4n)/2,
4Bn-12An=13n,
——》Bn={13n+12*[-(n^2+4n)/2]}/4=(-6n^2-11n)/4,
bn=Bn-B(n-1)=(-6n^2-11n)/4-[-6(n-1)^2-11(n-1)]/4=-12n-5/4,
2、cn=(1/2)^(2bn-3an)=2^(3n-2)=2*8^(n-1)
That is to say, {CN} is an equal ratio sequence with the first term C1 = 2 and the common ratio q = 8,
Cn=c1*（1-q^n)/(1-q)=2(8^n-1)/7.

It is known that the sum of the first n terms of the sequence {an} is Sn, and Sn = 2an-2 (n = 1,2,3...), in the sequence {BN}, B1 = 1, and the point (BN, BN + 1) is on the straight line X-Y + 2 = 0
(1) Finding the general terms an and BN of sequence {an} {BN}
(2) Note Sn = A1B1 + a2b2 +. + anbn, find SN

(1) From Sn = 2an-2, when n = 1, S1 = A1 = 2a1-2
The solution is A1 = 2 and sn-1 = 2an-1-2
Then sn-sn-1 = an = 2an-2 - (2an-1-2) = 2an-2an-1
It can be concluded that an = 2, an-1 is an equal ratio sequence, and the common ratio is q = 2
So an = A1 &; QN-1 = 2 &; 2N-1 = 2n, n ∈ n+
Because the point (BN, BN + 1) is on the line X-Y + 2 = 0,
Then bn-bn + 1 + 2 = 0, that is BN + 1-bn = 2
This sequence is an arithmetic sequence with tolerance d = 2 and B1 = 1
So BN = B1 + (n-1) d = 1 + (n-1) & _; 2 = 2N-1
When n = 1, B1 = 1, then BN = 2, n-1, n ∈ n+
(2) from (1), we can see that an &; BN = 2n &; (2n-1) = 2n + 1 &; n-2N
So:
　　Sn = 22•1—2+23•2—22 +24•3—23+… +2n•(n—1) —2n-1+2n+1•n—2n
　　=23•1+24•2+… +2n•(n—2) +2n+1•n—2 ①
　　2Sn=24•1+25•2+… +2n+1•(n—2) +2n+2•n—4 ②
Formula 1 - Formula 2
　　—Sn = 23+24+25+… +2n +2n+1•2—2n+2•n+2
　　=2+(23—23•2n-2)／(1—2) +2n+2•(1—n)
　　=—6—2n+1•(2 n—3)
In conclusion, Sn = 6 + 2n + 1 &; (2n-3) is solved
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