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Let Sn be the sum of the first n terms of the sequence an, Sn = kn * 2 + N, n ∈ n *, where k is a constant, then find A1, an
Sn=kn*2+n Sn-1=k(n-1)*2+n-1
an=Sn-Sn-1=k(2n-1)+1
a1=k+1
Given the first n terms of sequence {an} and Sn = - 1 / 2n ^ 2 + kn, K ∈ n *, and the maximum value of Sn is 8, find the constant K, find an? Using sn-sn-1 formula
Sn = - 1 / 2n ^ 2 + kn = - 1 / 2 (n ^ 2-2kn) = - 1 / 2 (n-k) ^ 2 + (1 / 2) k ^ 2, so the maximum value (1 / 2) k ^ 2 = 8 is obtained when n = K. because K ∈ n *, k = 4, so Sn = - 1 / 2n ^ 2 + kn = - 1 / 2n ^ 2 + 4nan = sn-sn-1 = = - 1 / 2n ^ 2 + 4N - [- 1 / 2 (n-1) ^ 2 + 4 (n-1)] = - N + 9 / 2
Let Sn be the sum of the first n terms of the sequence {an}, Sn = kn + N, n ∈ n *, where k is a constant
(1) find A1 and an; (2) if any m ∈ n *, am, a 2m, a 4m are equal proportion sequence, find the value of K
(1) When n = 1, A1 = S1 = K + 1; when n ≥ 2, an = SN-S (n-1) = kn + n-k (n-1) - (n-1) = 2Kn + (1-k) when n = 1, A1 = K + 1 is suitable for the above formula. An = 2Kn + (1-k) (2) am, A2M, A4M are in equal proportion to the sequence ∧ (A2M) = am * A4M ∧ [2K * 2m + (1-k)] = [2km + (1-k)]] [2km + (1-k)] [2km * 4m + (1-k)]; 16km + 8
Given the first n terms of sequence {an} and Sn = - 1 / 2n ^ 2 + kn, K ∈ n *, and the maximum value of Sn is 8.1), the constant k is determined
If K is an integer, it will get 4, but this is only a special case. If K is not an integer, what is the general method to solve this problem? For example, Sn = - 1 / 2n ^ 2 + 7 / 3N, when n = 2, it will get the maximum,
If Sn = - 1 / 2n ^ 2 + kn, and the maximum value of Sn is S2 (that is, n = 2 is the maximum)
How to get k = 7 / 3
S2 is - 1 / 2 (2) ^ 2 + (7 / 3) * 2
Sn = (- 1 / 2) n & # 178; + kn = (- 1 / 2) (n & # 178; - 2Kn + K & # 178;) + K & # 178; / 2 = (- 1 / 2) (n-k) &# 178; + K & # 178; / 2 if K is a natural number, then when n = k, Sn has the maximum value (SN) max = K & # 178; / 2 = 8K & # 178; = 16 k = 4. If K is not a natural number, then according to the actual situation: K0 and
Let an be an arithmetic sequence with non-zero tolerance, Sn be the sum of its first n terms, BN be an equal ratio sequence, and
a1=b1=2,S2=5b2,S4=25b3.
(1) Finding the general term formula an and BN of sequence an and BN
(2) Let CN satisfy CN = Sn * BN, and ask what is the maximum value of CN?
Let the general term formula of {a (n)} be: a (n) = 2 + D (n-1) {B (n)} be: B (n) = 2 × Q ^ (n-1), then the sum of the first n terms of {a (n)} is: S (n) = [a (1) + a (n)] n / 2 = [4 + D (n-1)] n / 2. According to the meaning of the problem, we can get the following solution: 4 + D (2-1)] × 2 / 2 = 5 × 2 × Q ^ (2-1) [4 + D (4-1)] × 4 / 2 = 25 × 2 × Q ^ (3-1)
The sum of the first n terms of the sequence {an} is Sn = 2an-2. The sequence {BN} is an arithmetic sequence with A1 as the first term and non-zero tolerance, and B1, B3 and B11 are proportional
(1) Finding the value of A1, A2, A3 (2) finding the general term formula of sequence {an} and {BN}
(1)S1=a1=2a1-2 -> a1=2
S2=a1+a2=2a2-2 -> a2=4
S3=a1+a2+a3=2a3-2 -> a3=8
(2)Sn=2an-2 ①
S(n-1)=2a(n-1)-2 ②
① (2) the equal ratio sequence of an = 2an-2a (n-1) - > an = 2A (n-1) - > {an} with the first term 2 and the common ratio 2 is obtained
So an = 2 * 2 ^ (n-1) = 2 ^ n
So BN = B1 + (n-1) d = 2 + (n-1) d
B1, B3, B11 are equal proportion sequence - > B3 ^ 2 = B1 * B11 - > (2 + 2D) ^ 2 = 2 * (2 + 10d) - > d ^ 2 = 3D - > d = 0 or D = 3
Because the tolerance is not zero, BN = 2 + 3 (n-1) = 3n-1
Let the sum of the first n terms of the arithmetic sequence {an} be Sn, and if the first terms A10 and S8 = S12, then the sum of n is equal when Sn reaches the minimum
S8=S12
So A9 + A10 + a11 + A12 = 0
a9+a12=a10+a11
So A10 + a11 = 0
So A10 = - a11
Again, a10
So A10 and a11 are both positive and negative
And A100
So when n = 10, Sn is the smallest
It is known that {an} is an arithmetic sequence, S10 > 0, S11
This arithmetic sequence is decreasing, S10 = (a1 + A10) * 10 / 2 = (A5 + A6) * 10 / 2 > 0, so A5 + A6 > 0, S11 = (a1 + a11) * 11 / 2 = (A6 + A6) * 11 / 2
It is known that the sum of the first n terms of sequence {an} is Sn = an2 + BN + C (C is not equal to 0)
It is known that the sum of the first n terms of the sequence {an} is Sn = an2 + BN + C (C is not equal to 0). This paper proves that {an} is an arithmetic sequence when n > = 2 and N belongs to n
It is proved that: A1 = S1 = a + B + CSN = an ^ 2 + BN + CS (n-1) = a (n-1) ^ 2 + B (n-1) + C = an ^ 2-2an + A + bn-b + C (when n ≥ 2) an = SN-S (n-1) = 2an-a + B, (when n ≥ 2) A1 = 2a-a + B = a + B, which is inconsistent with A1 = a + B + C. A (n + 1) = 2A (n + 1) - A + B a (n + 1) - an = 2A (when n ≥ 2), so when n ≥ 2
In sequence an, A1 = 5, an + 1 = an + 3, then the general term formula of this sequence is______ ．
∵ an + 1 = an + 3, ∵ an + 1-an = 3 ∵ sequence is an arithmetic sequence with the arithmetic as 3 and the first term as 5 ∵ an = 5 + 3 (n-1) = 3N + 2, so the answer is 3N + 2
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