It is proved that the sequence LG (1 + an) is an equal ratio sequence It's known

It is proved that the sequence LG (1 + an) is an equal ratio sequence It's known

A (n + 1) = [a (n)] ^ 2 + 2A (n), a (n + 1) + 1 = [a (n)] ^ 2 + 2A (n) + 1 = [a (n) + 1] ^ 2, if a (n + 1) + 1 = 0, then a (n) + 1 = 0,..., a (1) + 1 = 0, a (1) = - 1, a (n) = - 1. If a (1) is not - 1, then a (n) + 1 is not 0. Therefore, a (n + 1) + 1 = [a (n) + 1]

In known sequence {an}, A1 = 2, Anan + 1 + an + 1 = 2An
In known sequence {an}, A1 = 2, an * (an + 1) + (an + 1) = 2An, the general term formula for {an}

Solution:
an*a(n+1)+a(n+1)=2an
Divide both sides by an * (an + 1)
The results are as follows
1+1/an=2/a(n+1)
Let BN = 1 / an
Then: 2B (n + 1) = BN + 1
2[b(n+1)-1]=bn-1
[b(n+1)-1]/[bn-1]=1/2
Then: {bn-1} is an equal ratio sequence whose common ratio is 1 / 2
Then: bn-1 = (b1-1) * (1 / 2) ^ (n-1)
=(1/a1-1)*(1/2)^(n-1)
=-(1/2)^n
Then, BN = 1 - (1 / 2) ^ n
And BN = 1 / an
Then: an = 1 / [1 - (1 / 2) ^ n]
=[2^n]/[2^n-1]

Function f (x) = 2x + 3 / 3x, sequence an for n > = 2, n belongs to positive integer, there is always a general formula for finding an (an = f (1 / an-1), A1 = 1, 1)

f(x)=(2x+3)/3x
=2/3+1/x
So an = f [1 / a (n-1)]
=2/3+1/[1/a(n-1)]
=2/3+a(n-1)
an-a(n-1)=2/3
So an is the arithmetic sequence of D = 2 / 3
So an = a1 + (n-1) d
=1+2(n-1)/3
=2n/3+1/3

Given the function f (x) = x / x + 2, the sequence an satisfies A1 = 1, a (n + 1) = f (an) to find the general term formula an of sequence an