All the items of the arithmetic sequence an are positive, A1 = 3, the sum of the first n phases is Sn, the common ratio of BN is an arithmetic sequence of 2, and a3b3 = 56, b5s4 = 768 (1) Formula for finding the general term of an BN (2) Finding the first n terms and TN of {an · BN}

All the items of the arithmetic sequence an are positive, A1 = 3, the sum of the first n phases is Sn, the common ratio of BN is an arithmetic sequence of 2, and a3b3 = 56, b5s4 = 768 (1) Formula for finding the general term of an BN (2) Finding the first n terms and TN of {an · BN}

(1) if the tolerance is D and the common ratio is Q, then A3 = 3 + 2D, S4 = 12 + 6D, B5 = q ^ 2b3 = 4b3 (3 + 2D) * B3 = 56... ① 4b3 * (12 + 6D) = 768... ② divide ① to get d = 2 and substitute ① to get B3 = 8, that is, B1 = 2An = a1 + (n-1) d = 2n + 1bn = B1 * q ^ (n-1) = 2 ^ n (2) TN = 3 * 2 + 5 * 4 + 7 * 8 +... + (2n + 1) * 2 ^ n will make the general term common

It is known that {an} is an equal ratio sequence, A1 = 1, A5 = 256, Sn is the sum of the first n terms of {BN}, B1 = 2, 5s5 = 2s8
Finding the general formula of {an} and {BN}
Can the common ratio be - 4?

1) When {an} is an equal ratio sequence, A1 = 1, A5 = 256, let an = a1q ^ (n-1) = q ^ (n-1), = > q = (A5 / A1) ^ (1 / 4) = ± 4q = 4, an = 4 ^ (n-1); when Q = - 4, an = (- 4) ^ (n-1). 2) let BN = B1 + (n-1) d = 2 + (n-1) d, = > Sn = b1n + n (n-1) d / 2 = 2n + n (n-1) d / 2, = > S5 = 10 + 10d, S8 = 16 + 28d, 5s5 =

It is known that an is an equal ratio sequence of positive numbers, A1 = 1, A5 = 256, Sn is the sum of the first n terms of the equal difference sequence BN, B1 = 2, S3 = 15
1. The general formula of an and BN
2. Let CN = an × BN, find the first n terms and TN of the sequence {CN}

a1=1,a5=256,
a5=q^4=256
q=4
an=4^(n-1)
b1=2,S3＝3b1+3d=15
d=3
bn=2+3(n-1)=3n-1
So, an = 4 ^ (n-1), BN = 3n-1
2）cn＝an×bn＝（3n-1）4^(n-1）
Tn=c1+c2+.+cn
Tn=2*1+5*4+8*4^2+...+(3n-4)*4^(n-2）+（3n-1）4^(n-1）
4Tn= 2*4+5*4^2+...+(3n-7)*4^(n-2）+（3n-1）4^n
-3Tn=2+3*4+3*4^2+...+3*4^(n-2）-（3n-1）4^n
-3Tn=2-[4(1-4^(n-1）]-（3n-1）4^n
Tn=2/3+[n*4^n}