In the sequence an, A1 = in, a (n + 1) = 2An + 3n-4, in is a real number. For any in, it is proved that an is not an equal ratio sequence BN = a (n + 1) - an + 3, try to judge whether BN is an equal ratio sequence The formula of finding the same term of an

In the sequence an, A1 = in, a (n + 1) = 2An + 3n-4, in is a real number. For any in, it is proved that an is not an equal ratio sequence BN = a (n + 1) - an + 3, try to judge whether BN is an equal ratio sequence The formula of finding the same term of an

1) A1 = in, A2 = 2 in + 2, A3 = 4 in + 9, it is obvious that an cannot be an equal ratio sequence
2) BN = an + 3n-1; B (n + 1) = a (n + 1) + 3 (n + 1) - 1 = 2An + 3n-4 + 3N + 2 = 2 (an + 3n-1) = 2bn, so BN is an equal ratio sequence and the common ratio is 2
3) After calculating BN, we can find an on the other hand

In known sequence {an}, A1 = 1, an + a (n + 1) = 2 ^ n (n ∈ n *), BN = 3an
(1) Try to prove that the sequence {an-1 / 3 * 2 ^ n} is equal to the ratio sequence, and find the general term formula of the sequence {BN}
(2) In the sequence {BN}, is there a sequence of three consecutive equal difference terms? If so, find out all the terms that meet the conditions; if not, explain the reason

（1）
a(n+1)/2^(n+1)=-1/2*an/2^n+1/2
a(n+1)/2^(n+1)-1/3=-1/2[an/2^n-1/3]
So {an / 2 ^ n-1 / 3} is an equal ratio sequence, the first term is 1 / 6, and the common ratio is - 1 / 2
an=1/3[2^n-(-1)^(n-1)]
bn=2^n-(-1)^(n-1)
（2）
If it exists, let B (k-1), BK, B (K + 1) form an arithmetic sequence, when k is odd
There are B (k-1) = 2 ^ (k-1) + 1, BK = 2 ^ k-1, B (K + 1) = 2 ^ (K + 1) + 1,
2^(k+1)-2=2^(2k)+2^(k-1)+2^(k+1)+1
2^(2k)-2^(k-1)+3=0
If k = 1 and K should be greater than or equal to 2, then such K does not exist

In known sequence an, A1 = 3, a (n + 1) = 3an + 2, find SN

a（n+1）+1=3(an+1)
Let BN = an + 1, then BN is an equal ratio sequence with B1 = 4 and q = 3. TN = B1 (1-qn) / (1-Q) = 2x3 ^ n-2
And TN = Sn + N, so Sn = tn-n = 2x3 ^ n-n-2

It is known that the sum of the first n terms of {an} is Sn, A1 = 1, and 3an-1 + 2Sn = 3. Find the values of A1 and A2, and find the general term formula of the sequence {an}

3a1+2S2=3a1+2(a2+a1)=3×1+2（a2+1)=3,a2=-1
From 3an-1 + 2Sn = 3, Sn = 3 / 2 - (3 / 2) a (n-1)
an=sn-s(n-1)=3/2-（3/2）a (n-1)-[3/2-（3/2）a (n-2)]=(3/2)×[a(n-2)-a(n-1)]
2an=3a(n-2)-3a(n-1),2an-6a(n-1)=3a(n-2)-9a(n-1) ,an-3a(n-1)=(-3/2)[a(n-1)-3a(n-2)]
The sequence {an-3a (n-1)} is an equal ratio sequence, the common ratio = - 3 / 2, the first term = a2-3a1 = - 1-3 × 1 = - 4, and the number of terms is n-2,
∴an-3a(n-1)=-4×（-3/2）^( n-2-1)= 4×（3/2）^( n-3),
Sum of sequence {an-3a (n-1)} = - 4 × [1 - (- 3 / 2) ^ (n-2)] / [1 - (- 3 / 2)] = 2 / 5 + 2 / 5 × (3 / 2) ^ (n-2)
And the sum of sequence {an-3a (n-1)} = an-3a (n-1) + a (n-1) - 3A (n-2) + a (n-2) - 3A (n-3) + +a2-3a1
=an-a1-2[a(n-1)+a(n-2)+a(n-3)+…… +a2+a1]
=an-a1-2s(n-1)
=3an-a1-2sn=3an+3a(n-1)-a1-[3a(n-1)+2sn]=3an+3a(n-1)-1-3
=3[an+a(n-1)]-4
∴3[an+a(n-1)]-4=2/5+2/5×（3/2）^( n-2),
an+a(n-1)-22/15=(2/15)×（3/2）^( n-2)=1/5+(1/5)×（3/2）^( n-3)
an+a(n-1)-1/5=(1/5)×（3/2）^( n-3)
The sequence {an + a (n-1) - 1 / 5} is an equal ratio sequence, the common ratio = 3 / 2, the first term = A2 + A1-1 / 5 = - 1 + 1-1 / 5 = - 1 / 5, and the number of terms is n-2,
[an+a(n-1)-1/5]-[a(n-1)+a(n-2)-1/5]+[a(n-2)+a(n-3)-1/5]-[a(n-3)+a(n-4)-1/5]+…… +[a3+a2-1/5]
-[a2+a1-1/5]=an-a1=an-1
=[an+a(n-1)-1/5]+[a(n-2)+a(n-3)-1/5]+…… +[a3+a2-1/5] -[a(n-1)+a(n-2)-1/5]-…… -[a2+a1-1/5]
=[an+a(n-1)-1/5]+[a(n-2)+a(n-3)-1/5]+…… +[a3+a2-1/5]-[a(n-1)+a(n-2)-1/5]-…… -[a2+a1-1/5]
=(1/5)×（3/2）^( n-3)+(1/5)×（3/2）^( n-5)+(1/5)×（3/2）^( n-7)+…… +(1/5)×（3/2）^2+(1/5)×（3/2）^0
-(1/5)×（3/2）^( n-4)-(1/5)×（3/2）^( n-6)-(1/5)×（3/2）^( n-8)-…… -(1/5)×（3/2）^3-(1/5)×（3/2）^1
-(1/5)×（3/2）^(-1)
In the above formula, all addends and subtractions are equal ratio series, common ratio = (3 / 2) ^ 2 = 9 / 4, and the first terms are (1 / 5) × (3 / 2) ^ 0 = 1 / 5,
The number of (1 / 5) × (3 / 2) ^ (- 1) = (1 / 5) × (3 / 2) is (n-2) / 2
Therefore, the above formula = an-1 = {(1 / 5) × [1 - (9 / 4) ^ (n-2) / 2] / [1 - (3 / 2)] - {(1 / 5) × (3 / 2) [1 - (9 / 4) ^ (n-2) / 2] / [1 - (3 / 2)]
=(1/5)[(9/4)^(n-2)/2-1]=(1/5)[(3/2)^(n-2)-1]=(1/5)(3/2)^(n-2)-1/5
So, an = (1 / 5) (3 / 2) ^ (n-2) - 1 / 5 + A1 = (1 / 5) (3 / 2) ^ (n-2) + 4 / 5
an=sn-s(n-1)=(3/2)×[a(n-2)-a(n-1)]