The sequence {an} satisfies an + 1 = 2An 0 ≤ an ≤ 1 / 2 an + 1 = 2an-1 1 / 2 ≤ an

a1=6/7 1/2

Given the function f (x) = (x-1) ^ 2, G (x) = 4 (x-1), the sequence an satisfies A1 = 2, and (an + 1-an) g (an) + F (an) = 0
Given the function f (x) = (x-1) ^ 2, G (x) = 4 (x-1), the sequence an satisfies A1 = 2, and (a (n + 1) - an) g (an) + F (an) = 0
(note that N and N + 1 are subscripts). BN = 3 / 4 (n + 2) (an-1) (n-1 is not subscript, n is), if T ^ m / b ^ m

An-1 = (3 / 4) ^ (n-1) an = (3 / 4) ^ (n-1) + 1sn = (1 - (3 / 4) ^ n) / (1-3 / 4) + NSN = 4-4 * (3 / 4) ^ n + n ≥

It is proved that there are infinitely many prime numbers Q such that 4q + 3 is prime

It's more reasonable to prove with contradiction or some number theory theorem

How to prove that prime number set is infinite set

Suppose the prime number is finite
There must be a maximum
Suppose the maximum prime number is p
Let n = 2 * 3 * 5 * 7 * *p+1
That is, multiply all prime numbers and add 1
It is obvious that n > P
So n is a composite number
Then n can be divided by at least one prime number
Singular, with 2,3,5 , P remove n
The results are all 1
So n is either prime or has a prime factor greater than P
But all of these contradict that P is the largest prime
So the assumption is wrong
So there are countless prime numbers
So the set of prime numbers is infinite

Let p be an odd prime, positive integers m, n satisfy M / N = 1 + 1 / 2 + 1 / 3.. + 1 / P-1, (m, n) = 1, and prove PIM

Proof: because P is an odd prime number, it satisfies 1

P is a prime for any positive integer A. is there always a positive integer m such that MP = a ~ (p-1) - 1? If so, please prove it briefly

Consider (P + 1) / 2 integers m2, where m is 0,1,..., (p-1) / 2. It is not difficult to see that the difference between any two of these integers i2-j2 = (I + J) (I-J) cannot be divisible by P (please think about why?), which shows that the remainder of these integers divided by P is different

It is proved that P > n ^ (1 / 3) and N / P are prime numbers

Counter proof: let n / P not be a prime, then n / P = N1 * N2, N1, N2 are all positive integers and N1 > = P, N2 > = P
So: n = P * N1 * N2 > = P ^ 3 is the contradiction of PN ^ 1 / 3
Therefore, if the hypothesis is not true, it can be proved

For any prime number P, we prove that there are infinitely many positive integers n such that P can be divided by (2 ^ N-N)

Fermat's theorem is proved by Euler's theorem in number theory, but Euler's theorem itself is troublesome, but Fermat's theorem has another simple proof method
For prime P and an arbitrary n (n cannot be divided by P), let:
n = c1 mod p
2n = c2 mod p
3n = c3 mod p
.
in = ci mod p
.
(p-1)n = c(p-1) mod p
Because n is not divisible by P and P is prime, {CI} is not equal to each other, because if x, y exist

It is proved that a ^ n + B ^ n can be divided by P, P = a + B P > N, P is prime, n is odd, a and B are positive integers

a^n + b^n
= (a+b)[a^(n-1) - a^(n-2)b + …… + (-1)^k*a^k*b^(n-1-k)+ …… +b^(n-1)]
So a ^ n + B ^ n can be divisible by P
(the condition that P is prime is superfluous.)

It is proved that a ^ N-B ^ n can be divided by P, P = a + B, P > N, P is prime, n is even, a and B are positive integers
It is proved that a ^ N-B ^ n can be divisible by P
P = a + B, P > N, P is prime, n is even. A, B are positive integers

Let n = 2K
Then a ^ n - B ^ n
= (a^2)^k - (b^2)^k
= (a^2 - b^2)[a^(2k-2) + a^(2k-4)b^2 + …… + b^(2k-2)]
P = a + B can divide a ^ 2-B ^ 2
So a ^ N-B ^ n can be divisible by P

The sequence {an} satisfies an + 1 = 2An 0 ≤ an ≤ 1 / 2 an + 1 = 2an-1 1 / 2 ≤ an