Sequence (14 15:6:39) For the arithmetic sequence with odd items, the sum of odd items is 44, and the sum of even items is 33. Find the number of items and the middle items of the sequence

Sequence (14 15:6:39) For the arithmetic sequence with odd items, the sum of odd items is 44, and the sum of even items is 33. Find the number of items and the middle items of the sequence

The middle term is 44-33 = 11
If the middle term is odd, then
There are 33 / 11 * 2 * 2 + 1 = 13 items in the sequence
If the middle term is even, then
There are 44 / 11 * 2 * 2-1 = 15 items in the sequence

Sequence (1 13:15:50)
In an n-storey building, each floor can call n people to hold a meeting. Now, one person on each floor is designated to hold a meeting on the k-th floor. In order to minimize the total walking distance of N meeting personnel up and down the stairs, how much should k be taken?

Set the meeting on the k-floor, set the distance between each two floors as h, and the distance from the people below the k-floor to the k-floor is (k-1) h, the distance from the first floor to the k-floor is (K-2) h, and the distance from the people below the k-floor to the k-floor is (k-1) H / 2. Similarly, the distance from the people above the k-floor to the k-floor is (k-1) h, and the distance from the people below the k-floor to the k-floor is (S2 = H (...)

The first term of sequence {an} is A1 = 1, the first n terms and Sn satisfy an = 2Sn & # 178; (2sn-1), (n ≥ 2) between an and an. (1) prove that sequence {1 / Sn} is an arithmetic sequence; (2) find the general term formula of sequence {an}

⑴an=2Sn^2/(2Sn-1)
2an*Sn-an=2Sn^2
Because an = SN-S (n-1) so,
2[Sn-S(n-1)]*Sn-Sn+S(n-1)=2Sn^2so,
S(n-1)-Sn=2S(n-1)*Sn
So [1 / Sn] - [1 / S (n-1)] = 2
The sequence {1 / Sn} is the arithmetic sequence; 2. The general term formula of the sequence {an}
⑵[1/Sn]-[1/S(n-1)]=2
[1/S(n-1)]-[1/S(n-2)]=2
.
.
.
[1/S2]-[1/S1]=2
All of them are added together to get: [1 / Sn] - [1 / S1] = 2
（n-1)S1=a1=1
So 1 / Sn = 2n-1sn = 1 / (2n-1)
In this paper, we introduce an = 2Sn ^ 2 / (2sn-1),
The result is: an = (- 2) / [(2n-3) * (2n-1)]

Insert n numbers between a and B to make them form an arithmetic sequence with a and B, then the tolerance of the sequence is ()
A. b−anB. b−an+1C. b+an+1D. b−an+2

Let A1 = a, then an + 2 = B, and then let its tolerance be D, then an + 2 = a1 + (n + 2-1) d, that is, B = a + (n + 1) d, so d = B − an + 1

1. In the arithmetic sequence {an}, Sn = n * n + 2n, if {BN} = 1 / an * a (n + 1), find the first n terms and TN of the sequence {BN}

Sn = n * n + 2n, then sn-1 = (n-1) ^ 2 + 2 (n-1) = n ^ 2-1
an=Sn-Sn-1=2n+1
bn=1/an*a（n+1）=1/(2n+1)(2n+3)=1/2[1/(2n+1)-1/(2n+3)]
Tn=b1+b2+b3+...+bn
=1/2[1/3 - 1/5 + 1/5- 1/7 + 1/7 - 1/9 +.1/(2n+1)-1/(2n+3)]
=1/2[1/3-1/(2n+3)]
=n/3(2n+3)

The sum of the first n terms of arithmetic sequence
A power management office is located at a power pole along a highway. It is known that a car transports three power poles from the power management office each time. If the distance between two adjacent power poles is 50 meters, the total journey of the car is 35.5 kilometers. The distance between the latter power pole and the power management office is 2450 meters. How far is the first power pole from the power management office? 2. How many power poles have been planted?

Let's suppose that there are N poles and the number of times is m, then it must be 3M = n. let's suppose that the distance between the first pole and the electric pipe station is x, and the journey of the M th bus is 2x + (m-1) * 50 * 2, then we can sum the m times to 35.5, and according to the title, it has 2450 = (n-1) * 50 + 2x, so we can solve the problem

In the known sequence {an}, n belongs to n *, an > 0, and the sum of the first n terms is Sn, satisfying Sn = an + 1 under 2 radical signs
To prove that an is an arithmetic sequence
Can you make it clear?

Because 2 √ s (n) = a (n) + 1 2 √ s (n + 1) = a (n + 1) + 1, the square subtraction 4 (s (n + 1) - S (n)) = [a (n + 1) + 1] ^ 2 - [a (n) + 1] ^ 24 · a (n + 1) = [a (n + 1)] ^ 2 + 2 · a (n + 1) - [a (n)] ^ 2-2 · a (n) [a (n + 1) + a (n)] · [a (n + 1) - A (n

Solve a problem of arithmetic sequence in senior one,
Right or wrong:
1. The sum of the first several terms of the sequence {an} is Sn = PN ^ 2 + QN. Where P and Q are constants, then the sequence must be an arithmetic sequence
2. The sum of the first several terms of the sequence {an} is Sn = PN ^ 2 + QN + R. where p, Q and R are constants and R is not equal to 0, then the sequence must be an arithmetic sequence
Please tell me clearly how to judge!
There is another way:
Find the value of S = 1 ^ 2-2 ^ 2 + 3 ^ 2-4 ^ 2 +... + 99 ^ 2-100 ^ 2.

1 pair an = sn-sn-1 = 2pn-p + QA1 = S1 = P + Q2 is not right an = sn-sn-1 = 2pn-p + QS1 = P + Q + R ≠ A1 = P + q3s = 1 ^ 2-2 ^ 2 + 3 ^ 2-4 ^ 2 +... + 99 ^ 2-100 ^ 2 = (- 1) (1 + 2) + (- 1) (3 + 4) + (- 1) (5 + 6) +... + (- 1) (99 + 100) = (- 1) (1 + 2 + 3 + 4 +... + 99 + 100) = - 5050

It is known that the tolerance of arithmetic sequence {an} is not zero, and A9 = 0, positive integers m and N are not equal +am = a1 + a2 + …… Can an be true? If so, find the relationship between M and N. if so, give the reason

a1+a2+…… +am = a1 + a2 + …… An can be established
Because A9 = 0
S9=S8+a9=S8+0=S8
Let m = 9, n = 8 or M = 8, n = 9

It is known that the product of the first three terms of the increasing sequence {an} is 64, and A2-1, A3-3 and a4-9 are equal difference sequences. (1) find the general term formula an of the sequence {an}; (2) let BN = n · an, find the first n terms and SN of the sequence {BN}

(1) Let the common ratio be Q. from the meaning of the question, we get: A2 = 4, ∵ 2 (A3-3) = A2-1 + a4-9, ∵ 2 (4q-3) = 3 + 4q2-9, and the solution is: q = 2 ∵ an = 2n (2) ∵ Sn = B1 + B2 + +bn=1×2+2×22+… +n×2n∴2Sn=1×22+2×23+… +By subtracting (n-1) × 2n + n × 2n + 1, Sn = - 2-22-23 - -2n...