A 1 = 1, a 5 = 8A 2, B N = an + N, find the sum of the first n terms of B n

A 1 = 1, a 5 = 8A 2, B N = an + N, find the sum of the first n terms of B n

Let the ratio of an equal ratio sequence be Q
a1=1,a2=a1q=q
a5=a1q^4=q^4
a5=8a2=8a1q=8q
∴q^4=8q
∴q=2 an=a1q^(n-1)=2^(n-1)
bn=an+n=2^(n-1)+n
sn=b1+b2+b3+...+bn
=(1+1)+(2+2)+(4+3)+...2^(n-1)+n
=[1+2+4+...+2^(n-1)]+(1+2+3+...+n)
=(1-2^n)/(1-2)+(1+n)n/2=(n+n^2)/2-1+2^n

In the sequence {an}, | A1 | = 1, A5 = - 8a2, A5 > A2, then an=______ ．

∵ in the equal ratio sequence {an}, | A1 | = 1, A5 = - 8a2, A5 ＞ A2, ∵ A1 = ± 1. Let the common ratio be q, when A1 = 1, Q4 = - 8q, the solution is q = - 2, then A5 = (- 2) 4 = 16, A2 = - 2, satisfy A5 > A2, ∵ an = (- 2) n − 1. When A1 = - 1, - Q4 = 8q, the solution is q = - 2, then A5 = (- 2) 4 = - 16, A2 = - (...)

It is known that the positive proportional sequence {an} is an increasing sequence, and satisfies the general term formula of a1 + A5 = 246, a2a4 = 729 (1)
(2) Let BN = anlog3an + 1 (n ∈ n *), the sum of the first n terms of BN is TN, and TN is calculated

Let the common ratio be q, and the sequence is an increasing sequence, Q > 1
The sequence is equal ratio sequence, A1 A5 = A2 A4 = 729, and a1 + A5 = 246. A1 and A5 are two parts of the equation x & # 178; - 246x + 729 = 0
(x-3)(x-243)=0
X = 3 or x = 243
The sequence is incremental, A5 > A1
a1=3 a5=243
a1/q5=q⁴=81
q>1 q=3
an=a1q^(n-1)=3×3^(n-1)=3ⁿ
The general formula of sequence {an} is an = 3 & # 8319;
bn=anlog3[a(n+1)]=(n+1)×3ⁿ
Tn=b1+b2+...+bn=2×3+3×3²+4×3³+...+(n+1)×3ⁿ
3Tn==2×3²+3×3³...+n×3ⁿ+(n+1)×3^(n+1)
Tn-3Tn=-2Tn=2×3+3²+3³+...+3ⁿ-(n+1)×3^(n+1)
=1+3+...+3ⁿ -(n+1)×3^(n+1) +2
=1×[3^(n+1)-1]/(3-1) -(n+1)×3^(n+1) +2
=-(2n+1)×3^(n+1)/2 + 3/2
Tn=(2n+1)×3^(n+1) /4 -3/4

(2014. The three module in Hexi District) set Sn as the first n term and the 8a2+a5=0 of the geometric sequence {an}, then S5S2 equals ().
A. 11B. 5C. -8D. -11

Let the common ratio of the equal ratio sequence {an} be q, (Q ≠ 0) from the meaning of the title, 8a2 + A5 = 8a1q + a1q4 = 0, the solution is q = - 2, so s5s2 = A1 (1 − Q5) 1 − QA1 (1 − Q2) 1 − q = 1 − q51 − Q2 = 1 − (− 2) 51 − (− 2) 2 = - 11, so D is selected