If the sum of the first n terms of the arithmetic sequence is Sn, A12 is negative 8, S9 is negative 9, then S16 is

a12=-8
s9=9 a5=-9 a5=-1
d=(-8+1)/7=-1
an=-n+4
a1=3 a16=-12
S16=(a1+a16)/2 *16= -72

Let Sn be the sum of the first n terms of the arithmetic sequence {an}, A12 = - 8, S9 = - 9, then S16=______ ．

If S9 = 12 (a1 + A9) × 9 = - 9, and a1 + A9 = 2a5, we can get, A5 = - 1. From the properties of arithmetic sequence, we can get, a1 + a16 = A5 + A12, then S16 = 12 (a1 + a16) × 16 = 12 (A5 + A12) × 16 = - 72

It is known that the sum of the first n terms of the sequence {an} is Sn, satisfying Sn = 2an-2n,
1. Find the formula of sequence an
two

Sn=2an-2n,
n=1 a1=2a1-2 a1=2
Sn=2an-2n,
S(n+1)=2a（n+1）-2（n+1）
Two form subtraction
S（n+1）-Sn= 2a（n+1）-2an-2
a（n+1）= 2a（n+1）-2an-2
a（n+1）=2an+2
[a(n+1)+2]=2[an+2]
{an + 2} is an equal ratio sequence with the first term of 3 and the common ratio of 2
an+2= 3*2^(n-1)
an=3*2^(n-1) -2

In the sequence an, if A1 = 1 / 2 and the first n terms and Sn = n ^ 2An, then an=

（1）
S2 = 2^2 * a2 = a1 + a2 = 1/2 + a2
a2 = 1/6
S3 = 3^2 * a3 = a1 + a2 + a3 = 1/2 + 1/6 + a3
a3 = 1/12
S4 = 4^2 * a4 = a1 + a2 + a3 + a4 = 1/2 + 1/6 + 1/12 + a4
a4 = 1/20
(2)
Guess that the general formula of {an} is an = 1 / [n (n + 1)]
Certificate:
When n = 2, there is
S2 = 2^2 * a2 = a1 + a2 = 1/2 + a2
a2 = 1/6 = 1/[2*(2+1)]
Suppose that when n = n, an = 1 / [n (n + 1)], Sn = n ^ 2 * an = n / (n + 1), then
When n = n + 1, there is
SN+1 = (N+1)^2 * aN+1 = a1 + a2 + …… + aN + aN+1 = N/(N+1) + aN+1
aN+1 = [N/(N+1)]/[(N+1)^2 - 1] = 1/[(N+1)(N+2)]
therefore
When n = n + 1, the formula holds
Therefore, for any n, there is an + 1 = 1 / [(n + 1) (n + 2)], and the proposition holds

Given that the sum of the first n terms of the sequence {an} is Sn, A1 = 1 / 2, and Sn = n ^ 2an-n (n-1), find an

In the case of n ≥ 2, we will be able to take the Sn (N-N (n-1) sn-1 (n-1) (n-1) (n-1) (n-1) (n-1) (n-1) (n-1) (n-1) (n-1) (n-1) (n-1) (n-1) (n-1) (n-1) (n-1) (n-1) (n-1) (n-1) (n-1) (n-1) (n-1) (n-1) (n-1) (n-1) (n-1) (n-1) (n-1) (n-1) (n-1) (n-1) (n-1) (n-1) (n-1) (n-1) (n-1) (n-1) (n-2) (n-1) (n-2) (n-1) (n-2) (n-2) (n-1) (n-1) (n-1) (n-2) (n-1) (n-2) (n-1) (n-1) (n-1) (n-1) (n-1) (n + 0 (n + 1) an - (n -...)

The sum of the first n terms of the sequence {an} is SN. It is known that A1 = 12, Sn = n2an − n (n − 1), n = 1, 2 Write out the recurrence relation between Sn and sn-1 (n ≥ 2), and find the expression of Sn about n

From Sn = n2an-n (n-1) (n ≥ 2), we can get: SN = N2 (sn-sn-1) - n (n-1), that is, (n2-1) sn-n2sn-1 = n (n-1), so n + 1nsn − NN − 1sn − 1 = 1 for n ≥ 2

Let the sum of the first n terms of the sequence {an} be Sn = 2an-2n, (I) find A1, A4 (II) prove that {an + 1-2an} is an equal ratio sequence; (III) find the general formula of {an}

(1) because A1 = S1, 2A1 = S1 + 2, so A1 = 2, S1 = 2, from 2An = Sn + 2n, know: 2An + 1 = Sn + 1 = Sn + 1 + 2n + 1 = an + 1 + 2n + 1 = an + 1 + 1 + Sn + 2n + 2n + 1, get an + 1 = Sn + 2n + 2n + 1, then A2 = S1 = S1 = S1 = S1 = S1 = S1 = S1 = S1 = 2 + 22 = 2 + 22 = 2 + 22 = 6, S2 = 8 = 2 = 2 = 2, S1 = 2 = 2 = 2 = 2 + 23 = 8 + 23 = 8 + 23 = 8 + 23 = 8 + 23 = 8 + 23 = 8 + 23 = 8 + 23 = 8 + 23 = 16, S2 = 24, S2 = 24, S2 = 24, S2 = 24, A4 = 24, and A4 = S3 + 24 = S3 + 24 = 4 + 24 = 4 = 4 = 4 + 24 = 40 from 2An = S3 + 24 + 24 from 2An = 3 + ratio sequence. (Ⅲ) an = (an)- 2an-1）+2（an-1-2an-2）+… +2n-2（a2-2a1）+2n-1a1=（n+1）•2n-1

Given that the sum of the first n terms of a sequence an is Sn, A1 = 1, Sn = 2An + 1, then Sn =?

Solution
∵a(n+1)=S(n+1)-Sn
Sn=2a(n+1)=2[S(n+1)-Sn]
3Sn=2S(n+1)
S(n+1)/Sn=3/2
S1=1
∴Sn=S1*(3/2)^(n-1)=(3/2)^(n-1) (n>=1)

In the sequence {an}, if A1 = 1, an + 1 = 2An + 3 (n ≥ 1), find the first n terms of an and Sn
That's it.
Wrong number. It's the sum of the first n terms of n * an

Due to:
a(n+1)=2an+3
The results are as follows:
[a(n+1)+3]=2(an+3)
[a(n+1)+3]/(an+3)=2
Then: {an + 3} is an equal ratio sequence with a common ratio of 2
Then:
an+3
=(a1+3)*2^(n-1)
=4*2^(n-1)
=2^2*2^(n-1)
=2^(n+1)
Then:
an=2^(n+1)-3
Then:
nan=n[2^(n+1)-3]
Then:
Sn=1*a1+2*a2+...+n*an
=1*[2^2-3]+2*[2^3-3]+...+n*[2^(n+1)-3]
=[1*2^2+2*2^3+...+n*2^(n+1)]-3(1+2+...+n)
=[1*2^2+2*2^3+...+n*2^(n+1)]-3n(n+1)/2
Let TN = 1 * 2 ^ 2 + 2 * 2 ^ 3 +... + n * 2 ^ (n + 1)
Then:
2Tn=1*2^3+2*2^4+...+(n-1)*2^(n+1)+n*2^(n+2)
By subtracting the two formulas, we get the following result:
-Tn=1*2^2+1*2^3+...1*2^(n+1)-n*2^(n+2)
Then:
Tn
=-[2^2+2^3+...+2^(n+1)]+n*2^(n+2)
=-[4*(1-2^n)/(1-2)]+n*2^(n+2)
=(n-1)*2^(n+2)+4
Then:
Sn=(n-1)*2^(n+2)-3n(n+1)/2+4

In the arithmetic sequence {an}, A1 ＞ 0, S4 = S9, then when Sn takes the maximum value, n=______ ．

4A1 + 4 × 32D = 9a1 + 9 × 82d, the solution is A1 = - 6D.  Sn = − 6dn + n (n + 1) D2 = d2n2 − 13d2n = D2 (n − 132) 2 − 1698d,  A1 > 0, d < 0,  when n = 6 or 7, the maximum value of Sn is - 1698d

If the sum of the first n terms of the arithmetic sequence is Sn, A12 is negative 8, S9 is negative 9, then S16 is