Final warm-up (2) 17 questions, the first N-term sum of sequence {an} is Sn, known A1 = 1. An + 1 = (n + 2) Sn / N (n ∈ n *) It is proved that: (1) the equal ratio sequence of sequence {Sn / N}. (2) Sn + 1 = 4An focuses on the second question,

Final warm-up (2) 17 questions, the first N-term sum of sequence {an} is Sn, known A1 = 1. An + 1 = (n + 2) Sn / N (n ∈ n *) It is proved that: (1) the equal ratio sequence of sequence {Sn / N}. (2) Sn + 1 = 4An focuses on the second question,

Certificate:
one
a(n+1)=S(n+1)-Sn=(n+2)Sn/n
nS(n+1)-nSn=(n+2)Sn
nS(n+1)=2(n+1)Sn
Divide both sides of the equation by n (n + 1)
S(n+1)/(n+1)=2(Sn/n)
[S (n + 1) / (n + 1)] / (Sn / N) = 2, is the fixed value
S 1 / 1 = a 1 / 1 = 1 / 1 = 1, sequence {Sn / N} is an equal ratio sequence with 1 as the first term and 2 as the common ratio
two
Sn/n=1×2^(n-1)=2^(n-1)
Sn=n×2^(n-1) S(n+1)=(n+1)×2ⁿ
When n ≥ 2, an = SN-S (n-1) = n × 2 ^ (n-1) - (n-1) × 2 ^ (n-2) = (n + 1) × 2 ^ (n-2)
When n = 1, A1 = (1 + 1) × 2 ^ (1-2) = 2 × (1 / 2) = 1, which also satisfies the general formula
The general term formula of sequence {an} is an = (n + 1) × 2 ^ (n-2)
4An = 4 × (n + 1) × 2 ^ (n-2) = (n + 1) × 2 & # 8319; = s (n + 1), the equation holds

If the sum of the first n terms of the arithmetic sequence {an} is Sn and 6s5-5s3 = 5, then A4=______ ．

∵ Sn = Na1 + 12n (n-1) d ∵ S5 = 5A1 + 10d, S3 = 3A1 + 3D ∵ 6s5-5s3 = 30a1 + 60d - (15a1 + 15d) = 15a1 + 45d = 15 (a1 + 3D) = 15a4 = 5, the solution is A4 = 13, so the answer is: 13

If M > 1, am-1 + am + 1-am2 = 0, s2m-1 = 38, then M is equal to ()
A. 38B. 20C. 10D. 9

According to the property of arithmetic sequence, we can get: am-1 + am + 1 = 2am, then am-1 + am + 1-am2 = am (2-AM) = 0, the solution is: am = 0 or am = 2, if am is equal to 0, it is obvious that s2m-1 = (2m − 1) (a1 + A2M − 1) 2 = (2m-1) am = 38 does not hold, so am = 2, s2m-1 = (2m-1) am = 4M-2 = 38, the solution is m = 10

Given the sum of the first n terms of {an} and Sn = n square-20n, find the sum of the first n terms of {an} and TN
Given the sum of the first n terms of {an} and Sn = n & # 178; - 20n, find the sum of the first n terms of {an} and TN

When n = 1, A1 = S1 = - 19, when n ≥ 2, an = SN-S (n-1) = (n & # 178; - 20n) - [(n-1) & # 178; - 20 (n-1)] = 2n-1-20 = 2n-21, so TN = | A1 | + | A2 | + | A3 | + +|a10|+|a11|+…… +|If n ≤ 10, an < 0, then | an | = 21-2n, TN = (21-2 × 1) + (21-2 × 2) + (21