Given the sequence {an}, where a1 = 10, and when n ≥ 2, an = 5an-1 / 6an-1 + 5, find the general term formula of the sequence {an} Given the sequence {an}, where a1 = 10, and when n ≥ 2, an = (5an-1) / (6an-1) + 5, find the common term of the sequence {an}

Given the sequence {an}, where a1 = 10, and when n ≥ 2, an = 5an-1 / 6an-1 + 5, find the general term formula of the sequence {an} Given the sequence {an}, where a1 = 10, and when n ≥ 2, an = (5an-1) / (6an-1) + 5, find the common term of the sequence {an}

Man, I didn't write the recursion formula clearly. Add a bracket

Let the first n terms of sequence {an} and Sn, and A1 = 1, Sn = 4A (n-1) + 2 (n ≥ 2)
(1) Let BN = a (n + 1) - 2An and prove that {BN} is an equal ratio sequence
(2) Let CN = an / 2 ^ n and prove that {CN} is an arithmetic sequence
Note: n + 1, n-1 are subscripts

1、a1=1,Sn=4a(n-1)+2
S(n-1)=4a(n-2)+2
an=4[a(n-1)-a(n-2)]
an-2a(n-1)=2*[a(n-1)-2a(n-2)]
[an-2a(n-1)]/[a(n-1)-2a(n-2)]=2
So BN / b (n-1) = [a (n + 1) - 2An] / [an-2a (n-1)] = 2
The ratio sequence is BN}
2、cn=an／2^n
cn-c(n-1)=an／2^n-a(n-1)／2^(n-1)=an-2a(n-1)/2^n
Because: [an-2a (n-1)] / [a (n-1) - 2A (n-2)] = 2
So: B (n-1) = an-2a (n-1) = B1 * 2 ^ (n-2)
b1=a2-2a1=5-2=3
So: cn-c (n-1) = an / 2 ^ n-a (n-1) / 2 ^ (n-1) = an-2a (n-1) / 2 ^ n = 3 * 2 ^ (n-2) / 2 ^ n = 3 / 4
So:
{CN} is an arithmetic sequence

The known sequence an satisfies A1 = 2, Sn = 4A (n + 1) + 2 (n = 2,3,4...)
Question 1 proves that sequence a (n + 1) - 2An is equal ratio sequence
2 prove that the n-th power of sequence an / 2 is equal difference sequence
The general term formula an and Sn of 3-ball sequence an

an=Sn-Sn-1=4(an-1-an-2)
Sequence {an + 1-2an} 4,8,16
an+1-2an
=Sn+1-Sn -2an
=4an-4an-1-2an
=2an-4an-1
=2（an-2an-1）
=8(an-1-an-2)-4an-1
=4（an-1-2an-2）
The sequence {an + 1-2an} is called equal ratio sequence, the first term is 4, and the common ratio is 2
Sequence {an / 2 to the nth power} 1,2,3,4
Mathematical induction
The nth power of BN = an / 2
Let BN = an / 2 to the nth power = n,
b1=a1/2=2/2=1
b2=a2/4=8/4=2
If bn-1 = an-1 / 2 to the power of n-1 = n-1, BN = an / 2 to the power of n = n
an-1=（n-1）×2^(n-1),an=n×2^n,
Then BN + 1 = an + 1 / 2 ^ (n + 1) = 4 (an-an-1) / 2 ^ (n + 1) = 2 & # 178; × [n × 2 ^ n - (n-1) × 2 ^ (n-1)] / 2 ^ (n + 1) = [2 ^ n + (n-1) × 2 ^ (n-1)] / 2 ^ (n-1) = 2 + (n-1) = n + 1
The sequence {an / 2 to the nth power} is called arithmetic sequence
The leading formula of {an} is the nth power of an = n × 2
Sn=4an-1+2=4[（n-1）×2^(n-1)]+2=[（n-1）×2^(n+1)]+2
Dazzled, big head, no longer give points, too meaningless

In the sequence an, A1 = 1, a (n) = a (n-1) / 2A (n-1) + 1 (n is greater than or equal to 2) holds for all non-zero natural numbers n, 2 whose a power-1 is greater than or equal to an

1 / a (n) = 2 + 1 / a (n-1), so 1 / a (n) is the arithmetic sequence with tolerance 2, the general term is B (n) = 1 / a (n) = 1 / (2n-1), n is all non-zero natural numbers 2 ^ n-1 and 1 / (2n-1), one is an increasing function, the other is a decreasing function, and the minimum value of the increasing function is 1 > = the maximum value of the decreasing function is 1, so A1 = 1, a (n) = a (n-1) / 2A (n-1) + 1

An is a sequence of increasing numbers. For natural number n, if an is equal to the square of N plus BN, then the value range of B is

An = n ^ 2 + BN = (n + B / 2) ^ 2 - B ^ 2 / 4 An is an increasing sequence, so - B / 2 ≤ 1, B ≥ - 2

The known sequence {an} satisfies: A1 = 1, a (n + 1) = 3an / (an + 3), an is not equal to 0. (n belongs to natural number) the general term formula of conjecture {an}

A (n + 1) = 3an / (an + 3), in the reverse
1/a(n+1)-1/an=1/3
1/an=1/3 n +2/3
an=3/(n+2)

In the sequence {an}, an = | n-k | + | n-2k |. For any positive integer n, if an is greater than or equal to A3 = A4, what is the value range of K?

|3-k|+|3-2k|=|4-k|+|4-2k|
When k ≤ 3 / 2, A3 ≠ A4; when 3 / 2 < K ≤ 2, k = 2, A3 = A4 = 2; when 2 < K ≤ 3, A3 ≠ A4; when 3 < K ≤ 4, k = 3, A3 = A4 = 3; when k > 4, A3 ≠ A4
2. Experience shows that when n ≤ 5, it is true
3. When n ≥ 6, an = 2n-6 > A3 or an = 2n-9 ≥ A3
K = 2 or 3

Given the sequence an, A1 = 5, A2 = 2, an = 2an-1 + 3an-2, n is greater than or equal to 3, can we find the general term, if so, what is it?

an+an-1=3(an-1+an-2）
Let BN = an + an-1, then bn-1 = an-1 + An-2
The first term of {BN} is A2 + A1, and the common ratio is 3
We obtain an + an-1 = 10 * (3 Λ n-2)
Let an = (3 Λ n-2) * 10 * CN, an-1 = (3 Λ n-3) * 10 * Cn-1
3cn+cn-1=3
By changing CN, Cn-1 into · x, the solution is x = 3 / 4, 3CN + cn-1-3 / 4 = 3-3 / 4, and 3 [(CN) - 9 / 4] = - [(Cn-1) - 9 / 4]
{(CN) - 9 / 4} first term (C1) - 9 / 4 common ratio is - 1 / 3 equal ratio sequence
The general term of de CN brings in an = (3 Λ n-2) * 10 * CN de · an

Given that the sequence {an} satisfies A1 = 2, an + 1-an = an + 1 * an, then A31 is equal to

If both sides divide an * an + 1, 1 / an-1 / an + 1 = 1
1 / an + 1-1 / an = - 1, so the sequence {1 / an} is an arithmetic sequence
1/an=1/a1+(-1)*(n-1)
1/a31=1/2+(-1)*30
1/a31=-59/2
a31=-2/59

Let the positive sequence an, A1 = 1, A2 = 2, and an = An-2 divided by an-1 (n is greater than or equal to three) find an
1,2, N, n-1, n-2 after a are all subscripts

an=a(n-2)/a(n-1)
Let xn = log (2) (an)
Take logarithm and change to xn = - x (n-1) + X (n-2)
Then we can use the characteristic equation to calculate