An is an arithmetic sequence, an = 13 + (n-1) 2. An = log2bn. It is proved that BN is an arithmetic sequence

An is an arithmetic sequence, an = 13 + (n-1) 2. An = log2bn. It is proved that BN is an arithmetic sequence

an=log2bn
So BN = 2 ^ an = 2 ^ [13 + 2 (n-1)]
Then B (n-1) = 2 ^ [13 + 2 (n-2)]
So BN △ B (n-1)
=2^{[13+2(n-1)]-[13+2(n-2)]]
=2²
=4
The division of two adjacent terms is a colonization
So BN is an equal ratio sequence

The known sequence {log2 (an)} is an arithmetic sequence with zero first term and one tolerance
1) Finding the general term formula of sequence {an}
Let {BN} = (3n-1) * an. Find B1 + B2 + B3 +... + BN

log2(an)=n-1,an=2^(n-1)
bn=(3n-1)*an=(3n-1)*2^(n-1)
B1 + B2 + B3 +. + BN = ∑ [3N · 2 ^ (n-1)] (1 to n) -∑ [2 ^ (n-1)] (1 to n)
=3 ∑ [n · 2 ^ (n-1)] (1 to n) - 2 ^ n + 1
Let Sn = ∑ [n · 2 ^ (n-1)] (1 to n), then 2Sn = ∑ [n · 2 ^ n] (1 to n)
Sn=2Sn-Sn=1·2^1+2·2^2+3·2^3+… +(n-1)·2^(n-1)+n·2^n
-1·2^0-2·2^1-3·2^2-… -n·2^(n-1)
=-[2^1+2^2+2^3+… +2^(n-1)]-1+n·2^n=-2^n+1-1+n·2^n=(n-1)·2^n
∴b1+b2+b3+.+bn=3(n-1)·2^n-2^n+1=(3n-2)·2^n+1

If {log2an} is an arithmetic sequence with tolerance of - 1 and S6 = 38, then A1=______ ．

∵ {log2an} is an arithmetic sequence with tolerance of - 1 ∵ log2an = log2a1-n + 1 ∵ an = 2log2a1 − n + 1 = A1 · 2 − n + 1 ∵ S6 = A1 (1 + 12 +...) +132) = A1 · 1 − 1261 − 12 = 38, so the answer is: 421