In sequence an, A1 = 2, an = 1 / 3A (n-1) - 2 (n is greater than or equal to 2), then a2013 is

In sequence an, A1 = 2, an = 1 / 3A (n-1) - 2 (n is greater than or equal to 2), then a2013 is

Find out the general formula an = 5x (1 / 3) ^ (n-1) - 3 and bring it into 2013

The first n terms and Sn of sequence {an} satisfy Sn = Nan + 2n ^ 2-2n (n ∈ positive integer), and the value of a100-a10 is

By substituting a (n) = SN-S (n-1) into the conditional expression, Sn = nsn-ns (n-1) + 2n (n-1) → ns (n-1) = (n-1) Sn + 2n (n-1) is divided by n (n-1) → s (n-1) = Sn + 2 --------... (n-1)... N so the sequence {s (n) / N} is an arithmetic sequence with tolerance of - 2, so s (n) / N = (A1 / 1) - 2 (n-1) = (a1-2n + 2)

The sum of the first n terms of the known arithmetic sequence {an} is SN. Given that A10 = 30, A20 = 50, find the general term an; if Sn and other 242, find n

Because
a10=a1+9d
a20=a1+19d
therefore
a20-a10=10d
d=2
a1=12
an=12+2(n-1)=10+2n
Sn=242
Sn=n(a1+an)/2
=n[2a1+(n-1)d]/2
=n(24+2n-2)/2=242
Solve the above equation and get n = 11, n = - 22
Obviously n > 0
So n = 11

The sum of the first n terms of the arithmetic sequence {an} is Sn, and A10 = 30, A20 = 50 (1) find the general term an (2) if Sn = 242 find n

The step length is (50-30) / 10 = 2, then A1 = 30 - (10-1) * 2 = 12
(1)an=a1+(n-1)*2=12+(n-1)*2
(2)a1+an=a2+a(n-1)=a3+a(n-2)=...
Sn=a1+a2+a3+...+an
=(a1+an)*(n/2) => (12+(12+(n-1)*2))*(n/2)=242
==>N = 11 or - 22 (rounding off)

In the sequence {an}, if the first n terms and Sn = N2 − 8N are known, then the value of A5 is ()
A. -63B. -15C. 1D. 5

Then A5 = s5-s4 = (25-40) - (- 16) = 1

It is known that the sum of the first n terms of the sequence {an} is Sn = n square-8n
(1) The general term formula of {an}
(2) Finding the general term formula of {an}
(3) The first n terms of {an} and Sn formula

(1)
A1 = 1 ^ 2-8 × 1 = - 7
By the condition Sn = n ^ 2-8n ①
s(n-1)=(n-1)^2-8(n-1)… ②
① (2) SN-S (n-1) = 2n-9
From an = SN-S (n-1)
So an = 2n-9, this formula is suitable for A1
So the general formula of {an} is 2n-9
(2)
N is an integer, 2n-9 < 0 when n ≤ 4, 2n-9 > 0 when n ≥ 5
Thus the general term formula of {an}
When n ≤ 4, | an | = 9-2n
When n ≥ 5, | an | = 2n-9;
(3)
When n ≤ 4
Each item is negative, so remove the absolute value and add a negative sign
So Sn = 8n-n ^ 2 (n ≤ 4)
When n ≥ 5,
Because S4 = a1 + A2 + a3 + A4 = | - 7 | + | - 5 | + | - 3 | + | - 1 | = 16
So Sn = S4 + [1 + 3 +... + (2n-9)] = (1 + 2n-9) (n-4) / 2 + 16 = n ^ 2-8n + 32
So when n ≤ 4, Sn = 8n-n ^ 2
When n ≥ 5, Sn = n ^ 2-8n + 32

In the known arithmetic sequence {an}, an = 33-3n, find the maximum value of Sn, calculate A1 = 30, d = - 3, but why must an > 0?

Because Sn = a1 + A2 + a3 +... + an, if an > 0, Sn increases, if an = 0, the maximum n. this problem is easy to find that the maximum n satisfying an > = 0 is 10 or 11
Then the sum formula of arithmetic sequence is used to calculate Sn = Na1 + [n (n-1) / 2] D. the result is the same, and the maximum value is 165

Given that the general term formula of arithmetic sequence {an} is an = 22-3n, then when n =, Sn gets the maximum value, the maximum value is?

The general formula of arithmetic sequence {an} is an = 22-3n
Then: the first term of the arithmetic sequence {an} is 19, and the tolerance is - 3
So Sn = n (41-3n) / 2
And because: an = SN-S (n-1)
When an > 0, Sn increases the sequence; when an 0
When n = 8, a8 = 22-24 = - 20, Sn increasing sequence; when n ≥ 8, an

In the sequence {an}, A1 = 1, √ an - √ an + 1 = √ Anan + 1, find the general term formula an

∵,√an -√an+1=√[anan+1 ]
Divide both sides by √ [ana (n + 1)]
∴(,√an -√an+1)/√[anan+1 ]=1
∴1/√a(n+1)-1/√an=1
{1 / √ an} is the arithmetic sequence, the tolerance is 1, and the first term is 1
∴1/√an=1+（n-1)=n
∴an=1/n²

In the one hour answer plus 30 minutes sequence {an}, A1 = 2, A2 = 3, and {an * an + 1} is an equal ratio sequence with 3 as the common ratio, denote BN = a2n-1 + A2N
In the sequence {an}, A1 = 2, A2 = 3, and {an * an + 1} is an equal ratio sequence with 3 as the common ratio. Note BN = a2n-1 + A2N, and prove that {BN} is an equal ratio sequence. (where an, an + 1, a2n-1, n, N + 1, 2N-1, 2n in A2N are subscripts)
Wait an hour, 30 minutes

Because {an * an + 1} is an equal ratio sequence with 3 as the common ratio, so (an + 1) / (an-1) = 3, the sequence is a mixed sequence of two equal ratio sequences, that is, the odd number term is an equal ratio sequence with 2 as the first term and 3 as the common ratio, and the even number term is an equal ratio sequence with 3 as the first term and 3 as the common ratio, so BN + 1 / BN = [(A2N + 1) + (A2N + 2)] / [(a2n-1)