If the first n terms of decreasing arithmetic sequence {an} and Sn satisfy S5 = S10, the value of n is () A. 10B. 7C. 9D. 7 or 8

If the first n terms of decreasing arithmetic sequence {an} and Sn satisfy S5 = S10, the value of n is () A. 10B. 7C. 9D. 7 or 8

∵ S5 = S10, ∵ s10-s5 = A6 + A7 + A8 + A9 + A10 = 0. According to the properties of the arithmetic sequence, a8 = 0 ∵ arithmetic sequence {an} decreases, ∵ d < 0, that is, a7 > 0, A9 < 0. According to the properties of the sum of the sequence, S7 = S8 is the largest Sn, so D

The first n terms of sequence an and Sn satisfy Sn = n ^ 2-8n + 1. If BN = | an |, find the general term formula of sequence {BN}

Sn=n^2-8n+1
sn-1=n^2-10n+10
Subtraction gives an = 2n-9
When n < 4, BN = 9-2n
When n > 4, BN = 2n-9

The general term formula of sequence an is known as an = 1 / (4N & # 178; - 1). If the sum of the first n terms of sequence an and Sn = 7 / 15, what is n
Why?

Sn=a1+a2+a3+…… +an
=1/3+1/15+1/35+…… 1/(4n^2-1)
=(1-2/3)+(2/3-3/5)+(3/5-4/7)+…… +n/(2n-1)-(n+1)/(2n+1)
=1-(n+1)/(2n+3)
=n/(2n+1)=7/15
The solution is n = 7

If the sequence an is an equal ratio sequence, and A1 = 2, q = 3, find Sn = 1 / A1A2 + 1 / a2a3 +. + 1 / ana (n + 1)

a2=a1q=6
bn=1/ana(n+1)
Then BN / b (n-1) = a (n-1) / a (n + 1) = 1 / Q & # 178; = 1 / 9
That is, B1 = 1 / 12
The common ratio of BN is 1 / 9
So Sn = B1 + +bn
=1/12*(1-1/9^n)/(1-1/9)
=3/32*(1-1/9^n)

Given that the first term A1 of sequence an is not equal to 0 and the tolerance D is not equal to 0, find Sn = 1. / A1A2 + 1 / a2a3 +. + 1 / ana (n + 1)

Because 1 / Anan + 1 = 1 / an * (an + D) = 1 / D [1 / an-1 / (an + D)] = 1 / D [1 / an-1 / an + 1]
So 1 / A1A2 + 1 / a2a3 + +1/anan+1
=1/d[1/a1-1/a2+1/a3-1/a4.1/an-1/an+1]
=1/d（1/a1-1/an+1）
=1/d*（an+1-a1）/a1an+1
=n/a1(a1+nd)
I wish you a happy study

The sequence {an} satisfies A1 = 1, an = an-1 (1 + 2An) (n-1 is the subscript). 1. Prove that {1 / an} is an arithmetic sequence; 2. If A1A2 + a2a3 +... + ana (n + 1)
>16/33

Your question is wrong. The subscript is n + 1, not n-1
a(n)=a(n+1)(1+2an)
a(n)=a(n+1)+2a(n)a(n-1)
Divide both sides by a (n) a (n-1)
1/a(n+1)=1/a(n)+2
1/a(n+1)-1/a(n)=2
(1) So {1 / an} is an arithmetic sequence
The first item 1 / A1 = 1, the tolerance is 2
The second problem is also incomplete. We can use the split term to find the sum
（2）1/an=1+2(n-1)=2n-1
an=1/(2n-1)
an*a(n+1)=1/(2n-1)(2n+1) =(1/2)[1/(2n-1)-1/(2n+1)]
a1a2+a2a3+...+ana(n+1)
=(1/2)[1-1/3+1/3-1/5+.+1/(2n-1)-1/(2n+1)]
=(1/2)[1-1/(2n+1)]
=n/(2n+1)
n/(2n+1)>16/33
33n>16(2n+1)
n>16
So the minimum value of n is 17

The sum of the first n terms of the sequence an is Sn, A1 = 1, a (n + 1) = 2Sn + 11) the general formula for finding an (2) the items of the arithmetic sequence BN are positive, and the sum of the first n terms is TN,

an=2S(n-1)+11 ①
a(n+1)=2Sn+11 ②
② - 1
a(n+1)-an=2an
a(n+1)=3an
a(n+1)/an=3
This is an equal ratio sequence with a common ratio of 3
an=3^(n-1)

Let the first n terms of sequence {an} and Sn = N2, then A8=______ ．

∵ an = sn-sn-1 (n ≥ 2), Sn = N2 ∵ A8 = s8-s7 = 64-49 = 15, so the answer is 15

If the sum of the first n terms of the sequence {an} is Sn = 2n & # 178;, then A8=

When n = 1, A1 = S1 = 2
When n ≥ 2, an = SN-S (n-1) = 4n-2, A1 satisfies an = 4n-2
So the general formula of sequence an is: an = 4n-2
a8=30

The sum of the first n terms of sequence {an} is Sn = 1 / 2n & # 178; - 3 / 2n, find the general term an, find A4, find A3 +... + A8

a1=S1=﹣1
an=Sn-S﹙n-1﹚(n≥2﹚
=n-2
A 1 is suitable for an = n-2 (n ≥ 2)
∴an=n-2
a4=2,
a3+...+a8
=1+2+…… +6=21