In the known sequence {an}, A0 = 2, A1 = 3, A2 = 6, and for n ≥ 3, there is an = (n + 4) an-1-4nan-2 + (4n-8) an-3. (I) let the sequence {BN} satisfy BN = an-nan-1, n ∈ n *, prove that the sequence {BN + 1-2bn} is an equal ratio sequence, and find the general term formula of the sequence {BN}; (II) note n × (n-1) × X 2 × 1 = n!, find the first n terms and Sn of the sequence {Nan}

In the known sequence {an}, A0 = 2, A1 = 3, A2 = 6, and for n ≥ 3, there is an = (n + 4) an-1-4nan-2 + (4n-8) an-3. (I) let the sequence {BN} satisfy BN = an-nan-1, n ∈ n *, prove that the sequence {BN + 1-2bn} is an equal ratio sequence, and find the general term formula of the sequence {BN}; (II) note n × (n-1) × X 2 × 1 = n!, find the first n terms and Sn of the sequence {Nan}

(I) & nbsp; prove: from the condition, an-nan-1 = 4 [an-1 - (n-1) An-2] - 4 [An-2 - (n-2) an-3], then an + 1 - (n + 1) an = 4 [an-nan-1] - 4 [an-1 - (n-1) An-2] So {BN + 1-2bn} is an equal ratio sequence with the first term of - 2 and the common ratio of 2 4 points b2-2b1 = - 2, so BN + 1-2bn = 2N-1 (b2-2b1) = - 2n. Divide both sides by 2n + 1, BN + 12n + 1 − bn2n = - 12 So bn2n = B12 − 12 (n − 1), BN = 2n (1 − N2) 8 points (Ⅱ) an-2n = nan-1-n2n-1 = n (an-1-2n-1), let CN = an-2n, then CN = ncn-1 •2•1•c1=n（n-1）•… •2•1．∴an=n（n-1）•… •2•1+2n． … 12 points Nan = n · n · (n-1) · •2•1+n2n=（n+1）!-n!+n•2n，∴Sn=（2!-1!）+（3!-2!）+… +（n+1）!-n!+（1×2+2×22+… +n×2n）．… Let TN = 1 × 2 + 2 × 22 + +Then 2tn = 1 × 22 + 2 × 23 + +(n-1) × 2n + n × 2n + 1 +2n-n×2n+1，Tn=（n-1）2n+1+2．∴S^＝(n+1)!+(n−1)2n+1+1．… 16 points

If an is equal difference sequence and A1, A3, a7 are equal ratio sequence, what is (a1 + a3) / (A2 + A4)?

∵ an into arithmetic sequence
∴a3=a1+2d,a7=a1+6d,
∵ A1, A3, a7, in equal proportion sequence
∴a3^2=a1×a7
∴(a1+2d) ^2=a1×(a1+6d)
{A1 = 2D (D ≠ 0) or D = 0
When D ≠ 0,
a1+a3=6d
a2+a4=8d
Then (a1 + a3) / (A2 + A4) = 3 / 4
When d = 0, it is meaningless

It is known that {an} is an equal ratio sequence, and Sn is the sum of its first n terms
A more general question

Find the first term A1 and the common ratio Q and substitute them into the formula
When Q ≠ 1
an=a1q^(n-1)
sn=a1(1-q^n)/(1-q)
When q = 1
an=a1
sn=na1

Please answer the math series of senior one in detail, thank you! (31 13:41:48)
Let {an} be an arithmetic sequence, and {BN} be an arithmetic sequence with positive items, and A1 = B1 = 1, A3 + B5 = 21, A5 + B3 = 13
(1) General term formula for {an}, {BN}
(2) Find the first n terms and Sn of sequence {an / BN}

A1 = B1 = 1A3 + B5 = 1 + 2D + Q ^ 4 = 21a5 + B3 = 1 + 4D + Q ^ 2 = 13, so d = (20-q ^ 4) / 2 = (12-q ^ 2) / 440-2q ^ 4 = 12-q ^ 22q ^ 4-q ^ 2-28 = 0 (Q ^ 2-4) (2q ^ 2 + 7) = 0q ^ 2 = 4 {BN} is a positive number, Q > 0q = 2D = (12-q ^ 2) / 4 = 2An = 2N-1, BN = 2 ^ (n-1) an / BN = (2n-1) / 2 ^ (n-1) 2